Monotone Sequences

Definition. A sequence is called

(a) strictly increasing if a1 < a2 < a3 < …< an < …(b) increasing if a1 ≤ a2 ≤ a3 ≤ … ≤ an ≤ …(c) strictly decreasing if a1 > a2 > a3 > …> an > …(d) decreasing if a1 ≥ a2 ≥ a3 ≥ … ≥ an ≥ …

{ }1

ann

∞

=

A sequence that is either increasing or decreasing is called monotone.

Strictly decreasing.

Strictly increasing.

Not monotone.

Tests for monotonicity.

(1) Take the difference between successive terms

(a) If an+1 − an > 0 Strictly Increasing(b) If an+1 − an < 0 Strictly Decreasing(c) If an+1 − an ≥ 0 Increasing(d) If an+1 − an ≤ 0 Decreasing

(2) Take the ratio of Successive terms

(a) If an+1/an > 1 Strictly Increasing(b) If an+1/an < 1 Strictly Decreasing(c) If an+1/an ≥ 1 Increasing(d) If an+1/an ≤ 1 Decreasing

Either test can be used, depending on the form of the terms of the sequence.

Example. Show that the sequence

is strictly increasing.

1 2 3, , , , ,

2 3 4 1n

n+… …

Method 1.

Thus the sequence is strictly increasing.

2 2 21 ( 1) ( 2) ( 2 1) ( 2 )2 1 ( 1)( 2) ( 1)( 2)

n n n n n n n n nn n n n n n

+ + − + + + − +− = =+ + + + + +

1 0( 1)( 2)n n

= >+ +

Example. Show that the sequence

is strictly increasing.

1 2 3, , , , ,

2 3 4 1n

n+… …

Method 2.

2 21 ( 1) 2 1/ 122 1 ( 2) 2

n n n n nn n n n n n

+ + + += = >+ + + +

Again the sequence is strictly increasing.

Example. Determine whether the sequence

is strictly increasing or strictly decreasing.

111n n

∞ − =

Solution. We use the method of successive differences.

1 1 1 11 11 1 1

a an n n n n n

− = − − − = − + + +

( )( ) ( )

1 1 01 1

n nn n n n

+ −= = >+ +

Thus the sequence is strictly increasing.

Example. Determine whether the sequence

is strictly increasing or strictly decreasing.

4 1 1

nn n

∞ − =

Solution. We use the method of successive differences.

1 ( 1)(4 1) (4 3)1 4 3 4 1 (4 1)(4 3)

n n n n n na an n n n n n

+ + − − +− = − = + + − − +

2 2(4 3 1) (4 3 ) 1 0(4 1)(4 3) (4 1)(4 3)

n n n nn n n n

+ − − + −= = <− + − +

Thus the sequence is strictly decreasing.

Example. Determine whether the sequence

is strictly increasing or strictly decreasing.

{ }21

n nn

∞−

=

Solution. We use the method of successive differences.

2 2 2 2( 1) ( 1) 2 01

a a n n n n n n n n nn n

− = + − + − − =− − − + =− < +

Thus the sequence is strictly decreasing.

Example. Determine whether the sequence

is strictly increasing or strictly decreasing.

2

11 2

n

n n

∞

=+

Solution. We use the method of successive ratios.

Thus the sequence is strictly increasing.

1 1 1 2 12 2 2 1 2 2 2/ / 11 1 1 2 11 2 1 2 1 2 2 2 2

n n n n n na an n n n n n n n

+ + + ++ += = = >+ + + ++ + + +

i

Example. Determine whether the sequence

is strictly increasing or strictly decreasing.

10(2 )! 1

n

n n

∞

=

Solution. We use the method of successive ratios.

Thus the sequence is strictly decreasing.

1 110 10 10 (2 )! 10/ / 11 (2 2)! (2 )! (2 2)! (2 1)(2 2)10

n n n na an n nn n n n n

+ += = = <

+ + + + +i

Example. Determine whether the sequence

is strictly increasing or strictly decreasing.

52 1( )2

n

nn

∞

=

Solution. We use the method of successive ratios.

Thus the sequence is strictly decreasing.

21 15 5 5 2 5/ / 11 2 2 2 2 1( 1) ( 1) 5 222 2

n n n na an n n nn n n

+ += = = <

+ ++ +i

Another possible way to show that a sequence is monotone is in the case where an = f(n) and f(x) is a function whose derivative is always positive or always negative.

Example. Determine whether the sequence

is strictly increasing or strictly decreasing.

131n n

∞ − =

Solution.

Thus the function f is strictly increasing, so the sequence is strictly increasing.

1 1Let ( ) 3 . Then ( ) 0.2

f x f xx x

′= − = >

Example. Determine whether the sequence

is strictly increasing or strictly decreasing.

{ }21

nnen

∞−=

Solution.

Thus the function f is strictly decreasing when x ≥ 1, so the sequence is strictly decreasing.

( )2 2 2 2Let ( ) . Then ( ) 2 1 2 0x x x xf x xe f x e xe e x− − − −′= = − = − <

when x ≥ 1.

Example. Determine whether the sequence

is strictly increasing or strictly decreasing.

{ }1tan1

nn

∞−=

Solution.

Thus the function f is strictly increasing when x ≥ 1, so the sequence is strictly increasing.

11Let ( ) tan ( ). Then ( ) 021

f x x f xx

− ′= = >+

when x ≥ 1.

Properties that hold eventually.

The first few terms of a sequence can be dropped without changing the limit. (The first few terms really means any finite number - thus you can drop the first 100 billion terms if you want, without changing the limit.)

Example. Show that the following sequences are eventually strictly increasing or eventually strictly decreasing.

17(a) 1

nn n

∞ + =

{ }5(b) 1

nn en

∞−=

Example. Show that the following sequences are eventually strictly increasing or eventually strictly decreasing.

17(a) 1

nn n

∞ + =

{ }5(b) 1

nn en

∞−=

17(a) Look at the function ( ) . Thenf x xx

= +

217 17( ) 1 0 when 5. 2 2

xf x xx x

−′ = − = > ≥

Thus this sequence is eventually strictly increasing, The first few terms are: 18, 10.5, 8.66, 8.25, 8.4, …

Example. Show that the following sequences are eventually strictly increasing or eventually strictly decreasing.

17(a) 1

nn n

∞ + =

{ }5(b) 1

nn en

∞−=

5(b) Look at the function ( ) . Thenxf x x e−=

4( ) (5 ) 0 when 6. xf x x x e x−′ = − < ≥

Thus this sequence is eventually strictly decreasing.

One of the most important things we need to be able to do is to see whether or not a sequence has a limit, even when we have little or no idea what that limit is.

This will be the usual situation when we get to infinite series.The definition of limit implies that we must know the limit Lbefore we prove that L is the limit, but if we look at certain types of sequences, we can show that a limit exists without knowing what it is.

Theorem. If a sequence is eventually increasing, then either(1) The sequence has an upper bound, that is a constant M so that an ≤ M for all n, in which case it has a limit, or

(2) The sequence diverges to + ∞, lim .ann

=+∞→∞

Theorem. If a sequence is eventually decreasing, then either(1) The sequence has a lower bound, that is a constant M so that M ≤ an for all n, in which case it has a limit, or

(2) The sequence diverges to − ∞, lim .ann

=−∞→∞

Completeness of the Real Numbers

Problem. Define a sequence recursively by

(a) Show that an < 2 for n ≥ 1.

(b) Show that

(c) Show that the sequence converges and find its limit.

1 12, 2 for 1nn

a a a n+

= = + ≥

12 2 (2 )(1 ) for 1nn n n

a a na a+ − = − + ≥

Solution. The first few terms are

1 2 32, 2 2, 2 2 2 ,a a a= = + = + + …

(a) If an < 2, then1

2 2 2 2.nna a

+= + < + =

Since a1 < 2, it follows that a2 < 2, which then implies that a3 < 2, and so on. Formally this is a proof by mathematical induction.

Problem. Define a sequence recursively by

(a) Show that an < 2 for n ≥ 1.

(b) Show that

(c) Show that the sequence converges and find its limit.

12 2 (2 )(1 ) for 1nn n n

a a na a+ − = − + ≥

Solution.

(b) 1

1

2 2 so

2 2 2( 2) ( 2)( 1) (2 )( 1)n n

n nn n n n n n

aaa a a a aa a a

+

+

= +

− =− − − =− − + = − +

Since this expression is positive for all n ≥ 1, the sequence is increasing.

1 12, 2 for 1nn

a a a n+

= = + ≥

Problem. Define a sequence recursively by

(a) Show that an < 2 for n ≥ 1.

(b) Show that

(c) Show that the sequence converges and find its limit.

12 2 (2 )(1 ) for 1nn n n

a a na a+ − = − + ≥

Solution.

(c) Since the sequence is increasing and has an upper bound, it must converge (to a limit L somewhere between 2 and ).2Passing to the limit in the definition of the sequence, we see that: 22 ; 2 ( 2)( 1) 0.L L L L L L= + − − = − + =

Since L cannot be −1, it must be 2.

1 12, 2 for 1nn

a a a n+

= = + ≥

1 2

There is no rational number whose square is 2. Assume first that any common factors of p and q have been cancelled. If (p/q)2 = 2, then p2 = 2q2 so p is even, say p = 2r. Then p2 = 4r2 = 2q2 so q2 = 2r2 and q is even. However this is impossible, since then p and q would have a common factor of 2.

The Completeness of the real numbers, as expressed in previous theorems, is key to being able to know that a limit exists, without knowing what it is. These results do not hold for the rational numbers as the following result shows:

This shows that there are sequences of rational numbers that aremonotone and bounded, such as successive decimal approximations to the square root of two, but that do not have alimit in the rational numbers