series, sequences and convergence concepts of primary

Series, Sequences and Convergence

Concepts of primary interest:

Convergence: absolute and conditional

Tests: ratio, comparison, alternating term

Radius of convergence

Geometric, Riemann zeta, harmonic series

Sample calculations:

Summing the geometric series

Demonstration of conditional divergence

Applications of convergence tests

Convergence of (1-x)-1 and ez

Tools of the Trade

Converting sums to integrals

Estimating remainders

Finite Summations

( 00

( )N

nn

n)f x a x x

=

= −∑ [SSC.1]

Our primary calculus procedures, differentiation and integration, are linear operations. The

derivative of a sum is the sum of derivatives and so on. In the case of a finite sum the operation on

the sum makes sense as long as it makes sense term by term. Infinite sums leave that question open.

Is the derivative of the sum the sum of the derivatives taken term-by-term? An infinite sum must be

convergent in order to make any sense. The sum must be strongly and nicely convergent to ensure

that the derivative of an infinite sum is the sum of derivatives taken term-by-term. The tests for

convergence and the types of convergence are discussed below.

Infinite Series – a useful example: the Taylor’s series

( 00

( ) nn

n)f x a x x

∞

=

= −∑ [SSC.2]

The Taylor’s series with remainder term also plays a role in the discussion of convergence. It

clarifies the conditions required in order that the infinite series can be adequately represented by a

Contact: [email protected]

finite number of terms.

( )00

0

1( ) ( , )!

sns

nss

xd f

0f x x xs dx=

⎛ ⎞= − +⎜ ⎟⎝ ⎠

∑ R x x [SSC.3]

where [ ] ( ) [

11

0 01 *1( , ) for some * ,

1 !|

nn

n xn

d f ]0R x x x x x x xn dx

++

+

⎛ ⎞= −⎜ ⎟⎜ ⎟+⎝ ⎠

∈ and hence:

[ ] ( ) [ ] ( )1 1

1 1min 0 0 max 01 1

1 1( , )1 ! 1 !

| |n n

n nnn n

d f d fx x R x x x xn dx n dx

+ ++ +

+ +

⎛ ⎞ ⎛ ⎞− ≤ ≤ −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

[SSC.4]

The maximum and minimum are to be found in the interval [x0, x ].

The Sweetened Tea Sequence: A certain professor has an 18 oz. cup that he uses to brew tea in a

microwave oven. He observes that the tea is best when it is warm and when it contains between 0.75

and 4/3 packets of sweetener per 18 oz. The tea itself is great at one bag per 18 oz and even better if it

is stronger. The tea cools more rapidly when the cup is partially filled than when it is fully filled.

The professor concludes that he should brew an eighteen-ounce cup with one tea bag and one packet

of sweetener and then consume the tea until a fraction a remains. Water is then added up to the 18-

ounce level, and the tea is re-brewed after tossing in a new teabag and a packet of sweetener. The tea

bags are removed shortly after the tea is taken from the microwave oven.

What is the maximum fraction a of the tea that can remain before the tea is re-brewed each time if

the beverage is not to exceed the 4/3 packets per 18 ounce sweetener limit?

Let the 18 ounces be one unit of volume. Beginning with an empty cup, the first cup brewed has 1

packet. A fraction a of that first packet remains when the second cup is brewed with an additional

packet. The second cup has (1 + a) packets. A fraction a of that (1 + a) packets remains when

another packet is added for the third brewing. The third cup contains 1 + a (1 + a) = 1 + a + a2

packets. The pattern is now clear. The nth cup contains 1 + a + a2 + … + an-1 packets.

The finite sum 1 + a + a2 + … + an-1 is easily evaluated using the algebraic identity:

[1 + a + a2 + … + an-1] [1 – a ] = 1 - an [SSC.5]

The sweetness of the nth cup is Sn = (1 - an)/(1 – a), and the limiting sweetness of the tea is 1/(1 – a) in

packets per eighteen ounces.

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-2

The problem was to find the maximum value of a such that the sweetness is limited to be less than 4/3 packets per eighteen ounces.

1/(1 – a) < 4/3 if a < 1/4.

Exercise: In an effort to reduce the amount of sweetener consumed, the professor decides to add

only ½ packet of sweetener each brewing. He can tolerate this deprivation as long as he believes that

the sweetness level will reach 0.75 packets per 18 ounces eventually. How many ounces of each cup

should be consumed before re-brewing if the 0.75 sweetness level is to be reached ‘eventually’?

[12 oz.]

The Geometric Series (a.k.a. The Sweetened Tea Sequence)

The infinite series 2 3

0

1 ... ... (1 )i k

k

a a a a a a∞

1−

=

+ + + + + + = = −∑ is the geometric series.

The geometric series is absolutely convergent for |a| < 1.

Infinite Series and Sequences:

An infinite sum of scalar terms such as 0

i

i

a∞

−

=∑ is an infinite series. A sequence is an indexed list of

scalar values such as {b0, b1, b2, …. , }. For each infinite series, there is a sequence of particular

interest, the sequence of partial sums.

{ …. , Sn , ….} = {a0, a0 + a1, 2

0i

i

a=∑ , … ,

0

n

ii

a=∑ , … }

The sequence has a limit Slim if, for every ε, there exists an integer N such that | Slim – Sn| < ε for all

n > N. The infinite series converges and has a sum if the sequence of its partial sums has a limit. Its

sum is then that limit. When Slim exists, it will be represented as S, the limit of the sequence of partial

sums. The remainder after n terms Rn is (S – Sn ) = 1

ii n

a∞

= +∑ .

The symbol σ represents the sum of a series for which convergence has not been established.

Note that σ is not really defined if the series does not converge.


The infinite series converges only if R0

ii

a∞

=∑ N = .

1

0asii N

a N∞

= +

→ →∑ ∞

For a series to converge, the sequence of partial sums must approach a limit requiring at minimum

that the terms approach zero as the index grows large.

The convergence of one family of infinite series is to be studied before the question of convergence

is developed systematically. The Riemann Zeta function is defined as a series:

( )1

1( ) pk

p kζ∞

=

= ∑ . [SSC.6]

It is the tail-end (large index) terms that must be studied to establish the convergence properties. RN-1

= ( )1p

k N k

∞

=∑ . Does this sum approach zero as N becomes very large? The strategy is to sandwich the

value of the summed series between the values of the two integrals.

Georg Friedrich Bernhard Riemann (1826 - 1866)

(pronounced REE mahn) was a German mathematician who made important contributions to analysis and differential geometry, some of them paving the way for the later development of general relativity. Riemann was arguably the most influential mathematician of the middle of the nineteenth century. His published works are a small volume only, but opened up research areas combining analysis with geometry. These would subsequently be major parts of the theories of Riemannian geometry, algebraic geometry and complex manifold theory. The theory of Riemann surfaces was elaborated by Felix Klein and particularly Adolf Hurwitz. This area of mathematics was foundational in topology, and in the twenty-first century is still being applied in novel ways to mathematical physics. http://en.wikipedia.org/wiki/G._F._B._Riemann

Read the Tools of the Trade section on converting sums to integrals and then carefully prepare a

sketch to establish that:

( ) 11

ppNk N

dx dxk pNx x

∞∞

−=

< <∑∫∞

∫ [SSC.7]

The anti-derivatives are: [ ] ( )1 11 forln( ) for 1p

pdx p x px x p

− −⎧ − ≠⎪→ ⎨=⎪⎩

∫1

For p < 1, both integrals diverge so the sum must diverge.


http://en.wikipedia.org/wiki/Image:Georg_Friedrich_Bernhard_Riemann.jpeg

For p = 1, ( )1ln( ) ln( )pk N k

∞

=

∞ < < ∞∑ so the sum diverges.

For p > 1, ( ) ( ) 11

1 1( 1) ( 1) 1

p ppk N kp N p N

∞ 1−−

=

< <− − −

∑ so both bounds vanish in the large N limit.

The series converges.

Convergence of the Riemann Zeta Series:

( )1

converges for 11( )diverges for 1p

k

pp k p

ζ∞

=

>⎧= ⇒ ⎨ ≤⎩

∑

This integral technique is also used to determine bounds on the sum of a series. The first N terms are

summed, and the remainder term is bounded by a pair of integrals.

Riemann Zeta Series as a name is a little stuffy. In the future, the series ( )1

1p

k k

∞

=∑ is to be called the

p-series. The special case with p = 1 is to be called the harmonic series.

The series (1

1k

k∞

=∑ ) is called the harmonic series. It can be shown to diverge using the integral test leading to

comparison with the function ln(x) . The divergence, however, is very slow. Divergence of the harmonic series

was first demonstrated by Nicole d'Oresme (ca. 1323-1382), but was mislaid for several centuries (Havil

2003, p. 23; Derbyshire 2004, pp. 9-10). The result was proved again by Pietro Mengoli in 1647, by Johann

Bernoulli in 1687, and by Jakob Bernoulli shortly thereafter (Derbyshire 2004, pp. 9-10). http://mathworld.wolfram.com/HarmonicSeries.html

Divergent Series = Nonsense:

Begin with the series . One can subtract 3 from this series to find 1

3 3 9 27 81 ...n

n

σ∞

=

= = + + + +∑

2

3 3 9 27 81 ... 3n

n

σ σ∞

=

− = = + + + =∑ suggesting that 2 σ = - 3 or σ = -3/2. You can generate any

nonsense that you desire by treating a divergent series as meaningful.

How convergent is necessary?


A series is called conditionally convergent if, the series 1

kk

a∞

=∑ , as written, converges, but the series

1k

k

a∞

=∑ diverges. A reordering of a conditionally convergent series may converge to a different value

or fail to converge.

SEE: Eric W. Weisstein. "Riemann Series Theorem." From MathWorld--A Wolfram Web Resource.

http://mathworld.wolfram.com/RiemannSeriesTheorem.html

Consider the series: 1

1

( 1) 1 1 1 1 1 1 11 2 3 4 5 6 7

n

n n

+∞

=

−= − + − + − + −∑ . Note that the sum of the positive terms

appears to be greater than the sum of the absolute values of the negative terms although the

statement is meaningless since both sums diverge. Note that the sum of the negative terms is – (1/2)

ζ(1), a series that was demonstrated to diverge by comparison with integrations that bounded the

series sum. The conclusion is that the sum of the positive terms and the sum of the negative terms

are divergent. As noted by Boas, the sum can be arranged to sum to any value desired. Pick 1.5. One

sums values until the sum exceeds 1.5 (1 + 1/3 +1/5 ≈ 1.53) and then adds negative terms until the

sum drops below 1.5 (1 + 1/3 +1/5 - 1/2 ≈ 1.03). One then adds positive values until 1.5 is exceeded.

1 + 1/3 +1/5 - 1/2 + 1/7 + 1/9 +1/11 +1/13 +1/15 ≈ 1.52

This game can be continued forever as the sum of the positive terms and the sum of the negative

terms separately diverge.

1 + 1/3 +1/5 - 1/2 + 1/7 + 1/9 + 1/11 +1/13 +1/15 -1/4 ≈1.27

1+1/3+1/5-1/2 +1/7+1/9 +1/1 +1/13+1/15 -1/4+ 1/17 +1/19 +1/21 +1/23 +1/25 ≈1.51

Absolute Convergence: A series is absolutely convergent if the series that is the sum of the absolute

values of each term in the original series is convergent.

Rearranging the terms in an absolutely convergent series does not change the sum.

1 1

convergent absolutely convergentn nn n

a a∞ ∞

= =

⇒∑ ∑

All rearrangements yield the same sum.

An Infinite Series as a Function:


The Taylor’s series is a representation of a function as an infinite series. If the series converges in a

region, the function is said to be analytic in that region.

( ) (0

0 00 0

1( )!

nn n

n nn n x

d f )f x a x x x xn dx

∞ ∞

= =

= − = −∑ ∑ [SSC.8]

The series is convergent if, at each x, for every ε, there exists an Nε such that

( )0

00

1( )!

N nn

nn x

d ff x x xn dx

ε

ε=

− −∑ < [SSC.9]

If a single value of Nε can be found such that ( )

0

00

1( )!

N nn

nn x

d ff x x xn dx

ε

ε=

− − <∑ for all x in the

region, then the series is uniformly convergent in that region. If all the series involved are uniformly

convergent, then the series for the derivative of f(x) is the series formed by differentiating the series

for f(x) term by term, and the series for the integral of f(x) is the series formed by integrating the

series for f(x) term by term. These processes effectively interchange the order in which limits are

taken. Interchanging limiting processes requires uniform convergence to be guaranteed to make

sense. This property follows because a uniformly convergent series can effectively be replaced by its

first Nε terms and treated as a finite sum.

Radius of Convergence: (sometimes: circle of convergence)

Each series can be interpreted to be a function of a complex variable.

( ) (0

00 0

1( )!

nn

n nn n z

d f )0nf z a z z z z

n dz

∞ ∞

= =

= − = −∑ ∑ [SSC.10]

If the function is analytic, then there exists a value R, the radius of convergence, such that the

function converges at all points z for which 0z z R− < , at no points for which 0z z R− > . It may

converge at some points for which 0z z R− = and not for others. The value R can be infinity in

which case the function is an entire function and the series converges everywhere in the complex

plane.


Rzo

converges inside

diverges outside

A function that is analytic in the shaded region

will have a Taylor’s series expansion about z0

that converges at all points inside the circle (of

convergence) of radius R, at no points outside

that circle and that it may converge at some

points on the circle, but not at others. The

series converges uniformly and absolutely

inside any concentric circle with radius < R.

Consider the series ( )0

3s

sx∞

=∑ . It is the geometric series

0s

sa∞

=∑ which converges for |a| < 1. Thus the

series ( )0

3s

sx∞

=∑ has a radius of convergence of 3 centered on 0. The more general series

( )( 00s

)sb z z

∞

=

−∑ has a radius of convergence of b-1 centered on z0. These examples demonstrate that

a series with terms that are functions of a variable may converge for a restricted range of values of

the variable and diverge for values of the variable outside that range.

Exercise: How does the radius of convergence for ( )7

3s

sx∞

=∑ compare with that for ( )

03

s

sx∞

=∑ ?

The series expansion of (1 + x)-1 is 0

( )n

nx∞

=

−∑ . It follows from the ratio test, that the expansion has a

radius of convergence of 1 centered on x = 0. The sums for n = 0 to nmax = 50 are compared for x =

0.95 and 1.05.

Mathematica: converge = Table[{n,Sum[(-1)^m (95/100) ,{m,0,n}]},{n,0,50}] m

ListLinePlot[converge,PlotRange → {0,1}]

diverge = Table[{n,Sum[(-1)^m (105/100) ,{m,0,n}]},{n,0,50}]; m

ListLinePlot[diverge,PlotRange → All]


x = 0.95 convergent x = 1.05 divergent

x = 0.99 convergent x = 1.01 divergent

Convergence Criteria to this Point:

A series is just nonsense unless it converges so it is crucial that you learn to test for convergence.

What has been learned to this point?

Term Behavior: A series cannot converge unless |an| approaches zero for large n.

Geometric Series: 2 3

0

1 ... ... (1 )i k

k

a a a a a a∞

1−

=

+ + + + + + = = −∑ [SSC.11]

This series converges to (1 - a)-1 for |a| < 1 as demonstrated by brute force calculation. As the

geometric series is a positive term series, the convergence is absolute. The series diverges for |a| > 1.

Riemann Zeta Series: ( )1

converges for 11( )diverges for 1p

k

pp k p

ζ∞

=

>⎧= ⇒ ⎨ ≤⎩

∑ [SSC.12]

This family converges for p greater than 1. As they are a positive term series, this convergence is


absolute. In the case p =1, the zeta series is the harmonic series which is divergent and stands as the

boundary between converging and diverging Riemann zeta series.

Tests for Convergence

The Sanity Check: A series cannot converge unless |an| approaches zero for large n. The sanity

check is not a test for convergence; rather it provides a test to eliminate series that cannot converge.

The harmonic series is an example that passes the sanity check, but which fails to converge. That is:

passing the sanity check is necessary, but not sufficient.

The Integral Test: This test was presented by example in the Riemann zeta introduction. A finite

sum of defined values is always convergent. Trap the value of the remainder sum RN of a positive-

term series between two integrals with N dependent limits and show that both integrals approach

zero as N becomes large. Thus the remainder term goes to zero. This procedure is a special case of

the comparison test.

The Comparison Test for Convergence: Given that the series to test is 1

kk

a∞

=∑ , find an absolutely

convergent series such that |b1

kk

b∞

=∑ k| > |ak| for all k greater than some N. If such a series can be

found, the series 1

kk

a∞

=∑ is absolutely convergent, and the series

1k

k

a∞

=∑ converges to the same well-

defined value for all re-orderings. The greater than some N appears because convergence is not determined by any finite number of terms

in the series; it is determined by the tail-end behavior RN.

The Comparison Test for Divergence: Given that the series to test is 1

kk

a∞

=∑ , find a positive term

divergent series such that b1

kk

b∞

=∑ k < |ak| for all k greater than some N. If such a series can be found,

then the series 1

kk

a∞

=∑ diverges.


Corollaries of the Comparison Test:

i.) If is an absolutely convergent series, and 1

kk

b∞

=∑ n

n

ab tends to a finite limit,

then the series converges. 1

kk

a∞

=∑

ii.) If is a divergent positive-term series, and 1

kk

b∞

=∑ n

na

b tends to a finite

limit (non-zero), then the series 1

kk

a∞

=∑ diverges.

The Ratio Test: Given that the series to test is 1

kk

a∞

=∑ , form the sequence of the ratios of the next

term to the current term.

1

1 absolutely convergent1 the test fails!1 fails to converge

n

nn

aLimita

+

→∞

< ⇒⎧⎪→ =⎨⎪ > ⇒⎩

As the test involves the absolute values, the convergent case would be absolutely convergent. Note

that in the case that the limit is 1, the test fails. Some other test must be applied to determine the

convergence or divergence. The ratio test can be motivated by comparison test with the geometric

series in the cases a < 1 and > 1. In the case of the ratio approaching 1, the test fails. That is: no

definitive prediction can be made without additional information. One must appeal to more

specialized tests.

The ratio test is often applied just after the sanity check.

Alternating Term Series: If, after some finite number of terms (for all n greater than some finite N),

the terms in a series are alternately positive and negative, |an+1| < |an | and the , the series

converges. Note that the restriction |a

0nnLimit a

→∞=

n+1| < |an | obviates concerns about reordering the series.

More Convergence Tests: The convergence of series is an area of active interest, and there are

many specialized tests for convergence. The discussion of additional tests is to be appended to the

Tools of the Trade section just before the Problems as time permits or reason dictates.


Sample Calculations/Applications of Convergence Tests:

1.) The series 0

12 1m m

∞

= +∑ can be rewritten as 1

12 1m m

∞

= −∑ . This later positive term series dominates the series

1 1

1 1 12 2m mm m

∞ ∞

= =

=∑ ∑ term by term. As the final ‘p = 1’ series diverges, so must 0

12 1m m

∞

= +∑ by the comparison

test.

2.) The series 0 0

( 1)2 1m

m m

ma

m

∞ ∞

= =

−=

+∑ ∑ is an alternating term series. Clearly |am+1|< |am| for all m, and

. The series can be summed using a variety of tricks. In the Fourier series handout, the sum is

found to be

0mmLimit a

→∞=

04

( 1)2 1m

m

mπ

∞

=

−=

+∑ .

3.) The series 20 0

( 1)(2 1)m

m m

ma

m

∞ ∞

= =

−=

+∑ ∑ is an alternating term series. Clearly |am+1|< |am| for all m, and

. The series converges to a value defined to be Catalan’s constant (0.915965594… ). 0mmLimit a

→∞=

4.) The series 20 0

1(2 1)m

m m

am

∞ ∞

= =

=+∑ ∑ is an positive term series. Clearly, . 0mm

Limit a→∞

=

2 20 1

1 11(2 1) (2 1)m mm m

∞ ∞

= =

= ++ +∑ ∑ and 2

1

1(2 1)m m

∞

= +∑ is a positive term series that is dominated term by term

by 142

1 1

1 1(2 ) ( )m mm m

∞ ∞

= =

=∑ ∑ 2 . The final series, 21

1( )m m

∞

=∑ , converges, so all the series converge.

21

28

1(2 1)m m

π∞

=

=−∑ ; 2

1

26

1( )m m

π∞

=

=∑

5.) Discuss the convergence of the binomial expansion for the inverse of (1+x).

( ) 1

0 0set 1

( 1)( 2)...( 1) ( 1)( 1 1)( 1 2)...( )( 1)

! !1

0

s s s

s sn

n n n n s ss s

s

s

x x x∞ ∞

−

= =≡−

− − − + − − − − − −−+ = = =∑ ∑ x

∞

=∑

It is an alternating term series for positive x. As long as x > 0 the terms alternate in sign. For |x| < 1, |as+1|< |as|


for all s, and . The series converges by the alternating terms series test for 0 < x < 1. Taking a

different view, the series is a geometric series with a = - x so it converges for | - x | < 1 and hence for |x| < 1.

This result subsumes that found using the alternating term approach. This series is an expansion about zero so

it must converge for all z such that |z| < R and diverge for all |z| > R. The alternating term series test indicated

that R must be 1, and the geometric series result confirmed it.

0ssLimit a

→∞=

Exercise: Replace x by the complex value z. 0 0

( 1) ( )s s s

s szx

∞ ∞

= =

− −→∑ ∑ .

Let z = a i where a is real and 0 < a < 1. What is 0

( )s

sa i

∞

=

−∑ ? Take the limit a → 1. Compare the result with

½ (1 – i) = (1 + i)-1. Note than the series 0

( )s

si

∞

=

−∑ does not converge because |an| does not approach zero for

large n. Mathematica 6: Sum[(- I) ,{s,0,Infinity}]= 1/2- /2 s

Mathematica 5.2: Sum − I s, s, 0, Infinity@H L 8 <D Sum::div: Sum does not converge.= ‚s=0

∞

H− Ls

6.) Discuss the convergence of the binomial expansion for the inverse of (1-x).

( ) 1

0 0set 1

( 1)( 2)...( 1) ( 1)( 1 1)...( )! !

1 ( )0

( ) ss s

s sn

n n n n s s

s

xs s

x x∞ ∞

−

= =≡−

− − − + − − − −− = − = − =∑ ∑ x

∞

=∑

The sanity check is valid: . The term by term ratio is x so, by the ratio test, the series converges

absolutely as long as |x| < 1. Could the ratio test have been applied in the previous example? Note that the

ratio test result is valid in the complex plane so the series for (1-z)

0ssLimit a

→∞=

-1, converges everywhere inside a disk of

radius one centered on z = 0.

7.) Discuss the convergence of the Taylor’s series expansion of ez.

2 3

0 !1 ... ...

2 2(3) !

sz

s

szs

z z ze zs

∞

=

= = + + + + + +∑

The (s + 1) to s term-by-term ratio is z/(s + 1). The nth term in the sum has index s = n – 1.

The ratio test is:


1 1

1 convergent1 test fails! Here, 0for | |finite1 divergent

n n

n n sn n

n

a a zLimit Limit Limit za a

+ +

→∞ →∞ →∞

→∞

< ⇒⎧⎪→ = = →⎨⎪ > ⇒⎩

s

3

so, by the ratio test, the series converges absolutely for any finite |z|. A function with an unbounded or infinite

radius of convergence is well-defined and analytic over the entire complex plane and is called an entire

function.

8.) Discuss the convergence of the Taylor’s series expansion of [1 + 3x ] -1.

( )1 2

0

[1 3 ] 3 1 3 9 27 ....s

sx x x x x∞

−

=

+ = − = − + − +∑

The absolute value of the term-by-term ratio is ( )( )

133

3 3n

nx

xx x

+−−

= − = . The ratio test is:

1

13 | 3 | 1 !

1

n

n nn

convergentaLimit Limit x x test failsa

divergent

+

→∞ →∞

< ⇒⎧⎪= = → =⎨⎪ > ⇒⎩

so, by the ratio test, the series converges

absolutely for |x|< 1/3.

Tools of the Trade:

Converting Sums to Integrals

It is said that an integral is a sum of a huge number of small contributions, but some precision is

required before the statement becomes useful. Beginning with a function f(t) and a sequence of

values for t = {t1,t2,t3, ….,tN}, the sum 1

( )i N

ii

f t=

=∑ does not represent the integral ( )

t

tf t dt>

<∫ even if a

great many closely spaced values of t are used. Nothing has been included in the sum to represent

dt. One requires 1

( )i N

ii

if t=

=

Δ∑ t where ( ) [ ]1 11

2i it t it+ −Δ = − is the average interval between

sequential values of t values at ti. For well-behaved cases, the expression 1

( )i N

ii

if t t=

=

Δ∑ approaches the

Riemann sum definition of an integral as the t-axis is chopped up more and more finely. As


illustrated below, in the limit goes to zero, the sum tΔ1

( )i N

ii

if t t=

=

Δ∑ approaches the area under the

curve between t< and t>. That is: it represents ( )t

tf t dt>

<∫ provided the sequence of sums converges,

and life is good. The theory of integration is not the topic of this passage. The goal is simply to

remind you that the must be factored out of each term that is being summed in order to identify

the integrand.

tΔ

f(t)

t

t1 t2 ti tN

Δt

t< t>

f(t1)f(ti)

f(tN)

Δt

tk

f(tk)

area = f(tk) Δt

In the Fourier series handout discussion of the inner product, the function h(t) = g(t) f(t) was

considered at N equally spaced points between –T/2 and +T/2. This leads to the sum where

the N points

1

( )m N

mm

h t=

=∑

( ) ( ) ( )2 2mT T T

Nt = − − + Nm for (1 ≤ m ≤ N) have equal spacing Tt NΔ = and are

centered in each Δt wide interval. As the number of terms gets large, the sum must be divided by N

to ensure a result finite. This leads to (1

1( )m N

mm

Nh t=

=∑ ) . The rule for converting sums to integral

requires that Tt NΔ = be explicitly factored from each term in the sum. Thus

( ) ( ) ( ) ( )1 1 1

1 1 1( ) ( ) ( )][ ] [m N m N m N

m m mm m m

TN T N Th t h t h t t

= = =

= = =

→ = Δ∑ ∑ ∑ which becomes ( )/ 2

/ 21 ( )

T

T T f t dt+

−⎡ ⎤⎣ ⎦∫

as N gets large and Δt small.

Converting alternating term sums: ( )0

12 1 4

m

m mπ∞

=

−=

+∑ = 0.7853982


Anchor the evaluation. Sum through m = 7: ( )7

0

10.7542680

2 1

m

m m=

−=

+∑

The goal is to first estimate and then to provide bounds for the value of ( )8

12 1

m

m m

∞

=

−+∑ by

converting the remainder sum to an integral.

Converting a sum to an integral is sure to fail unless the change in the value of terms from one term

to the next is small compared to the absolute value of the terms. The conversion may fail anyway,

but unless this condition is met, it’s hopeless. The series above is an alternating (sign) term series.

The magnitude of the change in value of the sum from term to term is twice the magnitude of the

terms. A trick is needed. Combine adjacent positive and negative terms in pairs. It is a trick; think

about the sums below to verify that they are equal.

( )2

8 4 4

1 1 1 22 1 4 1 4 3 16 16 3

m

m m

∞ ∞ ∞

= = =

− ⎡ ⎤ ⎡ ⎤= − =⎢ ⎥ ⎢ ⎥+ + + + +⎣ ⎦ ⎣ ⎦∑ ∑ ∑

The last sum has the advantage that it is a sum of positive terms that vanish more rapidly than do the

alternating terms in the original series. It has better convergence properties.

To convert to an integral, must be factored from each term, but takes on every integer greater

than 4, the starting point, so .

Δ

1Δ =

2 2 24 4 4

2 2 2 1 19ln 0.27806416 16 3 16 16 3 16 16 3 4 17

d∞∞ ∞

= =

⎡ ⎤ ⎡ ⎤ ⎡ ⎤≡ Δ ≈ = ≈⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ + + + + +⎣ ⎦ ⎣ ⎦ ⎣ ⎦∑ ∑ ∫

( )7

20 4

1 2 0.78212 1 16 16 3

m

m

dm

∞

=

−+ ≈

+ + +∑ ∫

The exact value is π/4 ≈ 0.785398. With patience one can do better.

Exercise: Show that: (4 3

12

4 14

2 1 116 16 3 4 1 4 3

N

N N N

dud d∞ ∞ +

−

+

⎡ ⎤= − =⎢ ⎥+ + + +⎣ ⎦∫ ∫ ∫ )u . Use the result to

evaluate the integral above.

Bounding the sum: For suitable choices of limits, the integrals can provide both upper and lower


bounds for the sum. A careful sketch shows that for a monotone decreasing function F(n),

. 1

( ) ( ) ( )N N

n N

F x dx F n F x dx∞∞ ∞

−=

< <∑∫ ∫

The area of the shaded region is F(N), and F(x) as

drawn is a monotone decreasing function. It follows

that:

1

1( ) ( ) ( )

N N

N NF x dx F N F x dx

+

−< <∫ ∫

Exercise: Argue that, for a monotone decreasing

function,

1( ) ( ) ( )

N Nn N

F x dx F n F x dx∞∞ ∞

−=

< <∑∫ ∫

The shaded area is F(N) times 1. Make a sketch to

show that 1

1( ) ( ) ( )

N N

N NF x dx F N F x dx

+

−< <∫ ∫ given

that F(x) is monotone decreasing.

( ) ( ) ( )7 7

2 20 0 04 3

1 1 12 22 1 16 16 3 2 1 2 1 16 16 3

m m m

m m m

d dm m m

∞ ∞∞

= = =

− − −+ < < +

+ + + + + + +∑ ∑ ∑∫ ∫

or equivalently 0.7821 < ( )0

12 1

m

m m

∞

=

−+∑ < 0.7900. These results are consistent with a sum of π/4 ≈

0.7854. A current estimate might be chosen to be:

( )7

20 3.5

1 2 1 170.7542680 ln 0.78555872 1 16 16 3 4 15

m

m

dm

∞

=

− ⎛ ⎞ ⎛ ⎞+ ≈ + ≈⎜ ⎟ ⎜ ⎟+ + + ⎝ ⎠ ⎝ ⎠∑ ∫

It seems that splitting the difference is reasonable so long as one remembers the bounds on the

proven range.

Expansion/Identity Summary

Double Factorial: n!! = n(n-2)(n-4) … {terminating with 2 or 1.}

Function Series Summation Rconv


f(x) ( ) ( ) ( )0 00 0 0

22

2 ...1( ) ( ) 12!x xx x x x

df d ff x f xdx dx

− −⎛ ⎞= + + +⎜ ⎟⎝ ⎠

( )00

0

1!

ss

ss

xd f x x

s dx

∞

=

⎛ ⎞ −⎜ ⎟⎝ ⎠

∑ -

(x + y)n xny0 + n xn-1y1 + n(n-1)/2! xn-2y2 + n(n-1) (n-2)/3! xn-3y3 + ….. ( )

0

!( )! !

nn n s s

s

nn s sx y x −

=−+ = ∑ y -

sin(x) 3 5 7

3! 5! 7!x x xx − + − +… ( ) ( )

2 1

01

2 1

nn

n

xn

+∞

=

−+∑ !

∞

cos(x) 2 4 6

12! 4! 6!x x x

− + − +… ( ) ( )2

01

2 !

nn

n

xn

∞

=

−∑ ∞

tan(x) 3 5 7 72 17 62

3 15 315 2835x x x xx + + + + +… ½ π

ln[1+x] 2 3 4 5 6

2 3 4 5 6x x x x xx − + − + − +… ( ) 1

1

1n

nn xn

∞+

=

−∑ 1

(1+x)r 2 3( 1) ( 1)( 2)2! 3!1 ..r r r r rr x x x− − −+ + + +

1

( 1)( 2)...( 1)!1

s

r r r r ss

sx∞

=

− − − ++ ∑ 1

(1+x)n If n is a positive integer, the series terminates. finite sum ⇒ no convergence issue

ex2 3 4

12 3! 4!x x xx+ + + + +…

0 !

n

n

xn

∞

=∑ ∞

eix2 3 4 5

12 3! 4! 4!x x x xix i i+ − − + + −… ( ) ( ) ( )

2 2 1

01

2 ! 2 1 !

m mm

m

x xim n

+∞

=

⎛ ⎞− +⎜ ⎟⎜ ⎟+⎝ ⎠

∑

∞

Arcsin(x) 3 5 71 3 1 3 5

6 2 4 5 2 4 6 7x x xx + + − +

i i i i i …i i i i i

3 2

2

(2 1)!!6 (2 )!!(2 1)

m

m

x m xxm m

1+∞

=

−+ +

+∑ 1

Arctan(x) 3 5 7

3 5 7x x xx − + − +… ( ) ( )

2 1

01

2 1

nn

n

xn

+∞

=

−+∑ 1

Arctan(1/x) 3 5

1 1 12 3 5x x xπ

− + − +… ( )( ) 2 1

0

12 1

n

nn n x

π ∞

+=

−+

2 +∑ 1

cosh(x) 2 4 6

12! 4! 6!x x x

+ + + +… ( )2

0 2 !

n

n

xn

∞

=∑ ∞

sinh(x) 3 5 7

3! 5! 7!x x xx + + + +… ( )

2 1

0 2 1

n

n

xn

+∞

= +∑ ! ∞

tanh(x) 3 5 7 72 17 62

3 15 315 2835x x x xx − + − + −…

Taylor’s Series expansion of f(x) about the argument value xo:


( ) ( ) ( ) ( )0 00 0 0

22

020

0...

1 1( ) ( ) 12! !

ss

ss

x x xx x x xdf d f d ff x f x x xdx dx s dx

∞

=

− −⎛ ⎞ ⎛ ⎞= + + + = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑

Binomial Theorem: The theorem is an algebraic identity used to compute positive integer powers of

a binomial.

( ) ( ) ( ), ,0

! ( 1)( 2)...( 1);! ! !

nn n s s

n s n ss

ns

n n n n n sx y x yn s s s

β β−

=

− − − ++ = = = =

−∑

TO APPLY, convert to the form (1 + x)n where |x| < 1 and use the form:

2 3( 1) ( 1)( 2) ( 1)( 2)...( 1)2! 3! !(1 ) 1 ...n n n n n n n n n n s s

s xx nx x x− − − − − − +=+ + + + + +

Use the convention that only positive square roots considered for forms like 2 2a z+ as the

represent distances and that 2 2; ( ) ;...z z r z r z= − = − unless you have a reason to do

otherwise.

Euler's Identity: a relation between the exponential and the sinusoidal functions

also known as: The Euler-Lagrange identity cos sinie iθ θ θ= +

Problems

1.) Express the following series in summation form. Interpret the … as it just keeps on going.

a.) 1 1 1 11 ...3 5 7 9

− + − + − b.) 1 1 1 1 1 ...4 9 16 25 36

− + − + −

c.) 2 4 6

12 2 3 4 6!x x x

− + − +⋅ ⋅

… d.) 3 5 7

2 3 2 3 4 5 7!x x xx − + −⋅ ⋅ ⋅ ⋅

…+

2.) It is assumed that the index increments by one from term to term in each series. (Parts a.) and b.) demonstrate a technique to effectively increment by 2 or 4.) Show the explicit form of the first five terms of the following series:

a.) ( )8

12 1

m

m m

∞

=

−+∑ b.)

4

1 14 1 4 3

∞

=

⎡ ⎤−⎢ ⎥+ +⎣ ⎦∑

c.) 24

216 16 3

∞

=

⎡⎢ + +⎣ ⎦

∑ ⎤⎥ d.)

0

( 1)!

nn

n

xn

∞

=

−∑


e.) 1

21

( 1)n

n n

+∞

=

−∑ f.) 31

( 1)n

nn

∞

=

−∑

3.) Find the radius of convergence for the expansion:

( ) [ ]1

( 1)( 2)...( )!

11 1n s

s

n n n nx

ss

x∞

=

− − − −+ = + ∑

[ ]1 2 3 ( 1)( 2)...( 1 )( 1) ( 1)( 2)1 ...1! 2! 3! !

sn n n n sn n n n n nx x xs

− − − −− − −= + + + + + x

4.) The series expansion in x for ln(1+x) about x = 0 is:

( ) ( ) ( )1

1

ln 1 1ss

s

xx s∞

+

=

+ = −∑

Find the circle of convergence for this expansion. What happens to ( )ln 1 x+ for x = -1? Comment on

the circle of convergence in light of this observation.

5.) The series expansion in x for (1+x) -1 about x = 0 is:

( ) 1

0

( 1)1 s s

s

x x∞

−

=

−+ =∑

Find the radius of the circle of convergence for this expansion. What happens to (1+x) -1 for x = -1?

The radius of convergence for this problem is identical to that for ( ) 1

0

1 s

s

x x∞

−

=

− =∑ . Why?

Comment on the circle of convergence in light of this observation.

Series 1+x H−1L, x,0,7AH L 8 <E= 1−x+x2−x3+x4−x5+x6 −x7 +O x@ D8

6.) Use either of the sums 21

28

1(2 1)m m

π∞

=

=−∑ and 2

1

26

1( )m m

π∞

=

=∑ as a basis to establish the other.

[Hint: Which integers are involved in each sum? The sum of the odd terms plus the sum of the

even terms is the sum of all the terms. Relate the sum of the even terms to 21

1( )m m

∞

=∑ . ]

7.) For the series below, show 0ssLimit a

→∞= given that !s

sza s= . Given a value of z, what is the


minimum value of s (smin) that ensures that |as+1| < |as| for s > smin ? This problem is the basis for the

sanity check in the case of the series for 2

0 !1 ... ...

2 2(3) !

n sz

s

zs

z z ze zs

∞

=

= = + + + + + +∑ .

Find the radius of the circle of convergence of the series for ez.

8.) The power series solution method for: Legendre's Equation:

22

2(1 ) 2 ( 1) 0dPdx

d Px x Pdx

− − + + =

Leads to the series solutions:

Peven(x)

[ ][ ]2 40

22

0

6 ( 1) ( 1)( 1)1 ...(2)(1) 4!

ii

i

a x x

a x∞

=

⎧ ⎡ ⎤− + ++− − +⎪ ⎢ ⎥

⎪ ⎣ ⎦= ⎨⎪⎪⎩

∑

Podd(x)

[ ][ ]3 51

2 12 1

0

12) ( 1) ( 1)2 ( 1) ...(3)(2) 5!

ii

i

a x x x

a x∞

++

=

⎧ ⎡ ⎤− + +− ++ −⎪ ⎢ ⎥

⎪ ⎣ ⎦= ⎨⎪⎪⎩

∑

+

Use the recursion relation 2( 1) ( 1( 2)( 1)nn

n na an n+

)+ − +=+ +

and study the convergence of these series

as a function of x using the ratio test. The ratio test fails for x = ± 1. Other tests establish that the

general series diverges for x = ± 1. Finite results can be salvaged at x = ± 1 provided is restricted a

particular set of values. Given those values, the infinite series terminate after a finite number of

terms leading to a set of polynomials. Give the value spectrum for that leads to series truncated as

finite sums.

9.) It has been claimed that 21

26

1( )m m

π∞

=

=∑ . Show that 10

21

1( )m m=

∑ ≈ 1.54977. Use the integral method to

bound the value of R10 = 211

1( )m m

∞

=∑ and thereby demonstrate that 2

6π is a possible limit. Continue the

procedure to better bound the sum. Start with 100

21

1( )m m=

∑ ≈1.6349839002 and establish bounds for

R100. 2

6π ≈ 1.6449340.


10.) A ball is dropped from rest at a height H and falls with a uniform acceleration g downward

traveling a distance H to hit the floor the first time. (Ignore air resistance.) What is the time required

for the ball to fall the distance H? At that point the ball strikes a floor and rebounds upward with a

coefficient of restitution e. The speed of the ball just after the bounce is e times its speed just before

the bounce. What is the maximum height that the ball reaches on the first bounce? How long does it

take to reach that height? Find the total distance traveled and the total time elapsed during the first

bounce. Assuming the same coefficient of restitution for every bounce, construct sums to represent

the total distance that the ball travels and the total time elapsed since the ball was released. Sum the

series for an infinite number of bounces.

ANSWER: 2

2

1 2;1 1total total

e HD H Te g

⎡ ⎤+ +1 ee

⎡ ⎤= =⎢ ⎥ ⎢ ⎥− −⎣ ⎦⎣ ⎦

11.) Compute [cosh( )]ddx x and [sinh( )]d

dx x . Develop expressions for [cosh( )]nn

ddx x and [sinh( )]n

nddx x

for all positive integer values: n = 0, 1, 2, 3, … . Use these results, and generate the Taylor’s series

expansions about xo = 0 for hyperbolic cosine and hyperbolic sine. Find the radii of convergence for

hyperbolic cosine and hyperbolic sine.

12.) Parameter Calculus Used to Sum Series: Show that the series 0 0

n n

n n

ddan a a a

∞ ∞

= =

=∑ ∑ . Use this

to find . Extend the procedure to compute . 0

n

n

n a∞

=∑

0

( 1) n

n

n n a∞

=

−∑

13.) Convert a number in repetend form (eg. x = 2.191919... , a repeating decimal) into a ratio of

two integers. The decimal expression for x repeats every 2 digits. Form (102 x – x). The result is an

integer value. Hence x = (102 x – x)/ (102 – 1). is x expressed as a ratio of integers. The problem may be

attacked using series techniques as well. 2.191919… = 2 + 19 ( )1100

1m

m∞

=∑ . Use the geometric series


1

1 0

11

m m

m m aa aa a a

−

∞ ∞

= =− == =∑ ∑ 1− . So 2.1919…. = 198/99 + 19 (1/99) = 217/99. Employ this geometric series

approach to express 2.7128128128………… as a rational number, the ratio of integers.

14.) After making careful measurements of the electrostatic potential at point on the z axis for large

z, a student concluded that the potential could be expressed as: ( )104

mm

m

q azπε

∞

+=0

∑ . Sum the series to

show that this could be the potential of a single charge and identify its location.

References: 1. The Wolfram web site: mathworld.wolfram.com/ 2. K. F. Riley, M. P. Hobson and S. J. Bence, Mathematical Methods for Physics and Engineering, 2nd Ed., Cambridge, Cambridge UK (2002). 3. Donald A. McQuarrie, Mathematical Methods for Scientists and Engineers, University Science Books, Sausalito, CA (2003).


series, sequences and convergence concepts of primary

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