the comparison test let 0 a k b k for all k.. mika seppälä the comparison test comparison theorem...
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The Comparison Test
Let 0 ≤ ak ≤ bk for all k.
b
kk=1
∞
∑ converges ⇒ ak
k=1
∞
∑ converges.
Mika Seppälä
The Comparison Test
Comparison Theorem AAssume that 0 ≤ ak ≤ bk for all k.
If the series converges, then also the
series converges, and
a
kk=1
∞
∑ ≤ bk
k=1
∞
∑ .
b
kk=1
∞
∑
a
kk=1
∞
∑
Mika Seppälä
The Comparison Test
Claim
ak
k=1
∞
∑ converges.
Observe that the partial sums Sm = a1 + a2 + … + am form an increasing sequence since ak ≥ 0 for all k.
Proof
Mika Seppälä
The Comparison TEST
Claim
ak
k=1
∞
∑ converges.
The assumptions imply
b
kk=1
∞
∑
Observe that the sum is finite since this series converges.
Proof (cont’d)
Mika Seppälä
The Comparison test
Claim
ak
k=1
∞
∑ converges.
Proof (cont’d)
akk=1
m∑
The partial sums form a
bounded increasing sequence.
a
kk=1
∞
∑ = limm→ ∞
ak
k=1
m
∑
Hence the limit exists and is finite.
Mika Seppälä
The Comparison test
Comparison Theorem B
Assume that 0 ≤ ak ≤ bk for all k.
If the series diverges, then also the
series diverges.
ak
k=1
∞
∑
b
kk=1
∞
∑
Mika Seppälä
The Comparison Test
Claim
bk
k=1
∞
∑ diverges.
akk=1
∞∑
Since the series diverges, the
partial sums Sm form an unbounded set.
0 ≤S
m= a
kk=1
m
∑ ≤ bk
k=1
m
∑ .
The assumptions implyProof
Mika Seppälä
Claim
bk
k=1
∞
∑ diverges.
limm→ ∞
ak
k=1
m
∑ =∞.HenceProof (cont’d)
limm→ ∞
bkk=1
m∑ =∞.
Since , also
akk=1
m∑ ≤ bkk=1
m∑THE COMPARISON TEST
Mika Seppälä
Example
1
2k 2 + sink( )k=1
∞
∑Show that the series converges.
For all integer values of k, 1 < 2 + sin k < 3.
Solution
0 <1
3 ×2 k<
1
2 k 2 + sink( )<
12 k
Hence for all
integer values of k.
THE COMPARISON TEST
Mika Seppälä
Example
1
2k 2 + sink( )k=1
∞
∑Show that the series converges.
The series is a
convergent geometric series
Solution (cont’d)
1
2kk=1
∞∑
Hence converges by the
Comparison Theorem A.
1
2k 2 + sink( )k=1
∞
∑THE COMPARISON TEST
Mika Seppälä
Example
1
k 1−sin2 k( )k=1
∞
∑Show that the series diverges.
0 <1k
<1
k 1 −sin2 k( )
Hence for all positive
integer values of k.
For all positive integers k, Solution
0 <1 −sin2 k <1 .THE COMPARISON TEST
Mika Seppälä
Example
1
k 1−sin2 k( )k=1
∞
∑Show that the series diverges.
Since the Harmonic Seriesdiverges, also
Solution (cont’d)
diverges by the Comparison Theorem B.
1
kk=1
∞∑
1
k 1−sin2 k( )k=1
∞
∑
THE COMPARISON TEST
The Comparison Test
Let 0 ≤ ak ≤ bk for all k.
b
kk=1
∞
∑ converges ⇒ ak
k=1
∞
∑ converges.
a
kk=1
∞
∑ diverges ⇒ bk
k=1
∞
∑ diverges.