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Summary of Convergence Tests for Series and Solved ProblemsIntegral Test
Ratio TestRoot TestComparison Theorem for SeriesAlternating Series
Mika Seppälä: Series
Test Test Quantity
Converges if
Diverges if
Ratio q < 1 q > 1
Root r < 1 r > 1
Integral Int < ∞ Int = ∞
q =limk→ ∞
ak+1
ak
r =lim
k→ ∞a
kk
Int = f x( )dx
1
∞
∫
S = a
kk=1
∞
∑In the Integral Test we assume that there is a decreasing non-negative function f such that ak = f(k) for all k. The Test Quantity of the Integral Test is the improper integral of this function.
The above test quantities can be used to study the convergence of the series S.
Mika Seppälä: Series
Comparison Test and the Alternating Series Test
Comparison Test
bk
k=1
∞
∑ converges then also the series ak
k=1
∞
∑ converges, and ak
k=1
∞
∑ ≤ bk
k=1
∞
∑ .
Conversely, if ak
k=1
∞
∑ diverges, then also the series bk
k=1
∞
∑ diverges.
Assume that 0≤ ak ≤ bk for all k. If the series
The alternating series (−1)k+1akk=1
∞
∑ =a1 −a2 +a3 −K ,
converges if
Alternating Series Test
1
2 lim
k→ ∞ak =0.
∀k :ak ≥ak+1 ≥0, and
Mika Seppälä: Series
Error EstimatesError Estimate by the Integral Test
Let ak=f a
k( ) where f is a decreasing positive function. I f ak
k=1
∞
∑
converges by the Integral Test, then ak
k=1
M
∑ − ak
k=1
∞
∑ ≤ f x( )dxM
∞
∫ .
If the alternating series (−1)k+1ak1
∞∑ converges by the
Alternating Series Test, then (−1)k+1ak1
M∑ − (−1)k+1ak1
∞∑ ≤aM+1.
Error Estimates by the Alternating Series Test
Error of the approximation by the Mth partial sum.
This means that the error when estimating the sum of a converging alternating series is at most the absolute value of the first term left out.
Mika Seppälä: Series
Overview of Problems
2
Assume that ak and b
k are positive for all k, a
kk=1
∞
∑ converges
and that limk→ ∞
bk
ak
=L < ∞. Show that the series bk
k=1
∞
∑ converges.
1
Let ak and b
k be positive for all k. Assume that the series a
kk=1
∞
∑
and bk
k=1
∞
∑ both converge. Show that the series ak
k=1
∞
∑ bk converges.
3
We know that ak4k converges.
k=1
∞
∑ Do the following
series a) ak
−4( )k
k=1
∞
∑ and b) ak2 k
k=1
∞
∑ converge?
Mika Seppälä: Series
Overview of Problems
6
Compute 2
k k + 2( )k=1
∞
∑
7
k2
k3 +1k=1
∞
∑ 8
1
k3 + kk=1
∞
∑ 9
10 Estimate
1
k2
k=1
∞
∑ with error < 0 .001
Do the above series 4-5 and 7 – 9 converge or diverge?
Show that if a
k> 0 and lim
k→ ∞ka
k> 0, then a
kk=1
∞
∑ diverges.11
12 Show that if a
k> 0 and lim
k→ ∞k2a
k=1, then a
kk=1
∞
∑ converges.
4
1
k ln k( )k=2
∞
∑ 5
sin
1
k
⎛
⎝⎜
⎞
⎠⎟
k=1
∞
∑
Mika Seppälä: Series
Overview of Problems13
−1( )k+1
kk=1
∞
∑ 14
−1( )k+1
k
ln k( )k=1
∞
∑ 15
−1( )k+1
k + k( )
k − kk=2
∞
∑
16
17
Estimate
−1( )k+1
k3k=1
∞
∑ with error < 0 .001 .
18
Estimate −1( )
k
2k +1( )!k=0
∞
∑ with error < 0 .0001 .
19
−1( )k k2
2 kk=1
∞
∑ 20
−1( )k 2 k
k2k=1
∞
∑
Do the series in 13 – 16 converge?
Do the series in 19 – 20 converge?
Mika Seppälä: Series
Overview of Problems21
For which values of the parameter p the series
1
k pk=1
∞
∑ converges?
22
For which values of the parameter p the series 1
k lnp k( )k=2
∞
∑ converges?
23
k −22k + 3
⎛
⎝⎜
⎞
⎠⎟
k
k=1
∞
∑ 24
k20
k !k=1
∞
∑ 25
26
e−n n!n=1
∞
∑ 27
n2 e−n
n=1
∞
∑
28 Let F
n be the nth Fibonacci number.
Fn
2nn=1
∞
∑
29 Let a
1=1, a
n+1=
sin n( ) + cos n( )
na
n. a
nn=1
∞
∑
30 Show that the series
xn
n!
n=0
∞
∑ converges for all values of x.
Do the series given in Problems 23 – 29 converge?
Mika Seppälä: Series
Comparison Test
1
Let ak and b
k be positive for all k. Assume that
the series ak
k=1
∞
∑ and bk
k=1
∞
∑ both converge. Show
that the series ak
k=1
∞
∑ bk converges.Solution
Therefore there is a number m
1 such that k > m
1⇒ b
k< 1 .
Now k > m
1⇒ a
kb
k< a
k. Recall that by the assumptions a
kb
k> 0 .
Remark: it suffices to show that akb
km1
∞∑ converges because
akb
k1
∞∑ = akb
k1
m1 −1∑ + akb
km1
∞∑ and the sum akb
k1
m1 −1∑ is finite.
The Comparison Theorem now implies that a
kb
km1
∞∑ converges.
Mika Seppälä: Series
Comparison Test
2
Let ak and b
k be positive for all k. Assume that
the series ak
k=1
∞
∑ converges and that limk→ ∞
bk
ak
=L < ∞.
Show that the series bk
k=1
∞
∑ converges.Solution
Since lim
k→ ∞
bk
ak
=L, there is a number m1 such that k > m
1⇒
bk
ak
≤L +1 .
Therefore k > m
1⇒ b
k≤ L +1( )ak
.
Since a
k1
∞∑ converges, also L +1( )ak1
∞∑ converges.
Remark: it suffices to show that bkm
1
∞∑ converges because
bk1
∞∑ = bk1
m1 −1∑ + bkm1
∞∑ and the sum bk1
m1 −1∑ is finite.
The Comparison Theorem now implies that b
km1
∞∑ converges.
Mika Seppälä: Series
The Comparison Test
3
We know that ak4k converges.
k=1
∞
∑ Do the following
series a) ak
−4( )k
k=1
∞
∑ and b) ak2 k
k=1
∞
∑ converge?
Solution
I f ak= −1( )
k 1k4 k
then ak4 k
k=1
∞
∑ =−1( )
k
kk=1
∞
∑ converges by the
Alternating Series Test but ak
−4( )k
k=1
∞
∑ =1kk=1
∞
∑ is the Harmonic
Series which diverges.
The series a) needs not converge.
Example:
Mika Seppälä: Series
1
Hence the series
2 converges absolutely.kk
k k
a
The Comparison Test3
We know that ak4k converges.
k=1
∞
∑ Do the following
series a) ak
−4( )k
k=1
∞
∑ and b) ak2 k
k=1
∞
∑ converge?
Solution (cont’d)
Also k > k
1⇒ a
k2 k = a
k4 k 2 k
4 k
The series b) does converge.
Since ak4k
k=1
∞
∑ converges,
limk→ ∞
ak4 k =0.
The Geometric Series 1
2
⎛
⎝⎜
⎞
⎠⎟
k
k=k1
∞
∑ converges.
We conclude, by the Comparison Test,
that the series ak2k
k=1
∞
∑ converges.
<
12
⎛
⎝⎜
⎞
⎠⎟
k
.
<1
Mika Seppälä: Series
The Integral Test
4
Study the convergence of 1
k ln k( ).
k=2
∞
∑
Solution
The series 1
k ln k( )2
∞∑ diverges by the Integral Test
since dx
x ln x( )2
∞
∫ = limM→ ∞
dx
x ln x( )2
M
∫ = limM→ ∞
ln ln x( )( )⎤⎦2
M
= limM→ ∞
ln ln M( )( ) −ln ln 2( )( )( ) =∞.
Mika Seppälä: Series
Comparison Test
5 Study the convergence of sin
1
k
⎛
⎝⎜
⎞
⎠⎟ .
k=1
∞
∑
Solution We know that
x
2≤sin x( ) for 0 ≤x ≤1.
Therefore
1
2k≤sin
1k
⎛
⎝⎜
⎞
⎠⎟ for k ≥1.
The series
1
2k1
∞∑ diverges by the Integral Test
since dx
2x1
∞
∫ = limM→ ∞
dx2 x1
M
∫ = limM→ ∞
ln x( )
2
⎤
⎦
⎥⎥1
M
= limM→ ∞
ln M( )
2=∞.
The Comparison Theorem now implies that sin
1
k
⎛
⎝⎜
⎞
⎠⎟1
∞∑ diverges.
From Applications of Differentiation.
Mika Seppälä: Series
Partial Fraction Computation
6
Solution
Use the Partial Fraction Decomposition 2
k k + 2( )=1k−
1k + 2
.
Sm=
2
k k + 2( )k=1
m
∑ =1k−
1k + 2
⎛
⎝⎜
⎞
⎠⎟
k=1
m
∑
=
1kk=1
m
∑ −1
k + 2k=1
m
∑ = 1 +12+13+L +
1m
⎛
⎝⎜
⎞
⎠⎟ −
13+L +
1m
+1
m +1+
1m + 2
⎛
⎝⎜
⎞
⎠⎟
These terms cancel.
Mika Seppälä: Series
Comparison Test
7
Study the convergence of
k2
k3 +1.
k=1
∞
∑
Solution Observe that for k ≥1, 2k3 ≥k3 +1 .
Since the Harmonic Series 1
kk=1
∞
∑ diverges,
also the series 12kk=1
∞
∑ diverges.
By the Comparison Test, the series
k2
k3 +1k=1
∞
∑ diverges.
Mika Seppälä: Series
The Integral and the Comparison Tests
8
Study the convergence of
1
k3 + k.
k=1
∞
∑
Solution
Observe first that, by the Integral Test, 1
k3k=1
∞
∑ converges since
dx
x31
∞
∫ = limM→ ∞
dx
x31
M
∫ = limM→ ∞
−1
2 x21
M
= limM→ ∞
−1
2M2− −
12
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜
⎞
⎠⎟ =
12 converges.
Since 0 ≤1
k3 + k≤
1k3
, the series 1
k3 + kk=1
∞
∑ converges
by the Comparison Test.
Mika Seppälä: Series
The Integral Test
9
Study the convergence of 1
k ln k( ) ln ln k( )( ).
k=4
∞
∑
Solution
To use the Integral Test we have to compute
dx
x ln x( ) ln ln x( )( )4
∞
∫ = limM→ ∞
dx
x ln x( ) ln ln x( )( ).
4
M
∫
To compute dx
x ln x( ) ln ln x( )( )4
M
∫ use the substitution t =ln ln x( )( ).
Then dt =dx
x ln x( ) and we get
dx
x ln x( ) ln ln x( )( )4
M
∫ =dtt
ln ln 2( )( )
ln ln M( )( )
∫ =ln ln ln M( )( )( ) −ln ln ln 2( )( )( )M→ ∞ → ∞.
Hence the series diverges by the Integral Test.
Mika Seppälä: Series
The Integral Test10
Show that 1
k2
k=1
∞
∑ converges and
estimate the sum with error < 0 .001 Solution
The convergence follows from the fact that
dx
x21
∞
∫ = limM→ ∞
dx
x21
M
∫ = limM→ ∞
−1x
⎛
⎝⎜
⎞
⎠⎟
1
M
= limM→ ∞
−1M
− −1( )⎛
⎝⎜
⎞
⎠⎟ =1.
To find out how many terms we need to take in our approxiation,
determine M so that dx
x2M
∞
∫ < 0 .001 .
Computing as before,
dx
x2M
∞
∫ =1M
< 0 .001 if M > 1000 .
Computing 1000th partial sum by Maple we get the approximation 1.6439. The precise value of the above infinite sum is π2/6≈1.6449.
Mika Seppälä: Series
The Comparison Test
Show that if a
k> 0 and lim
k→ ∞ka
k> 0, then a
kk=1
∞
∑ diverges.
Solution Let lim
k→ ∞ka
k=b > 0 .
By the definition of limits: ∃m
1 such that k > m
1⇒ ka
k>
b2.
Hence, k > m
1⇒ a
k>
b2k
.
The series b
2kk=m1
∞
∑ , diverges since the Harmonic Series diverges.
Hence, by the Comparison Test, a
kk=1
∞
∑ diverges.
You can show this also directly by the Integral Test without referring to the Harmonic Series.
11
Mika Seppälä: Series
The Comparison Test12
Show that if a
k> 0 and lim
k→ ∞k2a
k=1, then a
kk=1
∞
∑ converges.
Solution Since lim
k→ ∞k2a
k=1, ∃m
1 such that k > m
1⇒ k2a
k< 2 .
Hence: k > m
1⇒ 0 < a
k<
2k2
.
The series 2
k2k=m1
∞
∑ converges by the Integral Test.
This follows since the improper integral
2dx
x2m
1
∞
∫ = limM→ ∞
2dx
x2m1
M
∫ = limM→ ∞
−2x
m1
M
= limM→ ∞
−2M
− −1m
1
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝⎜⎜
⎞
⎠⎟⎟=
1m
1
converges.
Hence, by the Comparison Test, a
kk=1
∞
∑ converges.
Mika Seppälä: Series
The Alternating Series Test13
Does the series
−1( )k+1
kk=1
∞
∑ converge?
Solution
The series is of the form −1( )k+1
ak
k=1
∞
∑
where ak=
1
k.
Hence the sequence a
k( ) is decreasing and limk→ ∞
ak=0.
Hence, by the Alternating Series Test,
−1( )
k+1
kk=1
∞
∑ converges.
Mika Seppälä: Series
The Alternating Series Test14
Does the series −1( )
k+1k
ln k( )k=1
∞
∑ converge?
Solution
The series is of the form −1( )k+1
ak
k=1
∞
∑
where ak=
k
ln k( ).
Since limk→ ∞
ak=lim
k→ ∞
k
ln k( )=∞,
the series −1( )
k+1k
ln k( )k=2
∞
∑ diverges.
Mika Seppälä: Series
The Alternating Series Test15
Does the series
−1( )k+1
k + k( )
k − kk=2
∞
∑ converge?
Solution
The series is of the form −1( )k+1
ak
k=2
∞
∑
where ak=
k + k
k − k.
Since limk→ ∞
ak=lim
k→ ∞
k + k
k − k=lim
k→ ∞
1 +1
k
1 −1
k
=1 ≠0,
the series −1( )
k+1k + k( )
k − kk=2
∞
∑ diverges.
Mika Seppälä: Series
The Alternating Series Test16
Does the series −1( )
ksin
πk
⎛
⎝⎜
⎞
⎠⎟
k=2
∞
∑ converge?
Solution
The series is of the form −1( )k
ak
k=2
∞
∑
where ak=sin
πk
⎛
⎝⎜
⎞
⎠⎟ > 0 for all k ≥2.
Since limk→ ∞
ak=0 and since the sequence a
k( ) is decreasing,
the series −1( )ksin
πk
⎛
⎝⎜
⎞
⎠⎟
k=2
∞
∑ converges
by the Alternating Series Test.
This follows from the fact that the sine function is increasing for 0≤x≤π/2.
Mika Seppälä: Series
The Alternating Series Test17
Estimate the sum
−1( )k+1
k3k=1
∞
∑ with error < 0 .001 .
Solution
The series is a convergent alternating seriesq
of the form −1( )k+1
ak
k=1
∞
∑ where ak=
1k3
.
The error of the approximation −1( )
k+1
k3k=1
m
∑ ≈−1( )
k+1
k3k=1
∞
∑
is ≤1
m +1( )3.
31
The error is 0.001 if 0.001, 1
i.e., if 1 10 or 9.
m
m m
Hence the desired estimate is
−1( )k+1
k3k=1
9
∑ ≈0.90211 .
Mika Seppälä: Series
The Alternating Series Test18
Estimate the sum −1( )
k
2k +1( )!k=0
∞
∑ with error < 0 .0001 .
Solution
The series is a convergent alternating series
of the form −1( )k
ak
k=0
∞
∑ where ak=
1
2k +1( )!.
The error of the approximation −1( )
k
2k +1( )!k=0
m
∑ ≈−1( )
k
2k +1( )!k=0
∞
∑
is ≤1
2m + 3( )!.
1The error is 0.0001 if 0.0001,
2 3 !
i.e., if 3.
m
m
Hence the desired estimate is −1( )
k+1
2k +1( )!k=0
3
∑ =42415040
≈0.841468 .
Mika Seppälä: Series
The Alternating Series Test19
Does the series −1( )
k k2
2 kk=1
∞
∑ converge?
Solution
The series is of the form −1( )k
ak
k=1
∞
∑
where ak=
k2
2 k.
Since limk→ ∞
ak=lim
k→ ∞
k2
2 k=0,
the series −1( )k k2
2 kk=1
∞
∑ converges.
Use l’Hospital’s Rule.
19
Mika Seppälä: Series
The Alternating Series Test20
Does the series −1( )
k 2 k
k2k=1
∞
∑ converge?
Solution
The series is of the form −1( )k
ak
k=1
∞
∑
where ak=2 k
k2.
Since limk→ ∞
ak=lim
k→ ∞
2 k
k2=∞,
the series −1( )k 2 k
k2k=1
∞
∑ diverges.
Use l’Hospital’s Rule.
Mika Seppälä: Series
The Integral Test21
For which values of the parameter p
the series 1
k pk=1
∞
∑ converges?
Solution
The improper integral
dx
xp1
∞
∫ converges if and only if p > 1 .
Hence, by the Integral Test, 1
k pk=1
∞
∑ converges
if and only if p > 1 .
Mika Seppälä: Series
The Integral Test22
For which values of the parameter p
the series 1
k lnp k( )k=2
∞
∑ converges?
Solution
Hence, by the Integral Test, 1
k lnp k( )k=1
∞
∑ converges
if and only if p > 1 .
To compute dx
x lnp x( )1
M
∫ use the substitution t =ln x( ) , dt =dxx
.
=t−p+1
1 −pln2
lnM
=lnM( )
1−p
1 −p−
ln2( )1−p
1 −pM→ ∞ → −
ln2( )1−p
1 −p if 1 −p < 0 .
This requires that p≠1. If p=1, the corresponding improper integral diverges.
Mika Seppälä: Series
The Root Test23
Does the series
k −22k + 3
⎛
⎝⎜
⎞
⎠⎟
k
k=1
∞
∑ converge?
Solution
ak
k =k −22k + 3
⎛
⎝⎜
⎞
⎠⎟
k
k =k −22k + 3
=1 −
2k
2 +3k
k→ ∞ → 12
< 1 .
Hence
k −22k + 3
⎛
⎝⎜
⎞
⎠⎟
k
k=1
∞
∑ converges by the Root Test.
Use the Root Test.
Mika Seppälä: Series
The Ratio Test24
Does the series
k20
k !k=1
∞
∑ converge?
Solution
ak+1
ak
=
k +1( )20
k +1( )!
k20
k !
=k +1
k
⎛
⎝⎜
⎞
⎠⎟
20k !
k +1( )!= 1 +
1k
⎛
⎝⎜
⎞
⎠⎟
201
k +1k→ ∞ → 0 .
Hence
k20
k !k=1
∞
∑ converges by the Ratio Test.
Use the Ratio Test.
Mika Seppälä: Series
The Comparison Test25
Does the series
cos k( ) + sin k( )
k2k=1
∞
∑ converge?
Solution
ak=
cos k( ) + sin k( )
k2<
2k2
for all k.
Use the Comparison Test.
Hence the series
cos k( ) + sin k( )
k2k=1
∞
∑ converges absolutely.
According to Problem 21.
Conclude that the series converges.
Mika Seppälä: Series
The Ratio Test26
Solution
an+1
an
=e
− n+1( ) n +1( )!
e−n n!=
en
en+1
n +1( )!
n!=1e
n +1( )n→ ∞ → ∞
Hence e−n n!
n=1
∞
∑ diverges by the Ratio Test.
Use the Ratio Test.
Mika Seppälä: Series
The Ratio Test27
Solution
an+1
an
=n +1( )
2e
− n+1( )
n2 e−n=
n +1n
⎛
⎝⎜
⎞
⎠⎟
2en
en+1= 1 +
1n
⎛
⎝⎜
⎞
⎠⎟
21e
n→ ∞ → 1e
Hence n2 e−n
n=1
∞
∑ converges.
Use the Ratio Test.
Mika Seppälä: Series
The Ratio Test
28
Let F0=0,F
1=1 and F
n=F
n−1+ F
n−2. The numbers F
n are the Fibonacci
numbers. They can be determined by the formula Fn=α n −βn
α −β where
α =1 + 5
2 and β =
1 − 52
.
Solution
Does the series
Fn
2nn=1
∞
∑ converge?
an+1
an
=
Fn+1
2 n+1
Fn
2 n
=F
n+1
Fn
2 n
2 n+1
=
α −βα
⎛
⎝⎜⎞
⎠⎟
n
β
1 −βα
⎛
⎝⎜⎞
⎠⎟
n
12
Conclude that the series converges by the Ratio Test.
=
α n+1 −βn+1
α −βα n −βn
α −β
12
=α n+1 −βn+1
α n −βn
12
n→ ∞ → α2
< 1 .
Now use the fact that β < α.
Hence 0 <βα
< 1 and limn→ ∞
βα
⎛
⎝⎜
⎞
⎠⎟
n
=0.
Mika Seppälä: Series
The Ratio Test
29
Let a1=1, a
n+1=
sin n( ) + cos n( )
na
n.
Does the series an
n=1
∞
∑ converge?
Solution
an+1
an
=
sin n( ) + cos n( )
na
n
an
=sin n( ) + cos n( )
nn→ ∞ → 0
The series converges by the Ratio Test.
Observe that for all positive integers n, sin(n) + cos(n) ≠0. Hence, for every n, an≠ 0, and the above ratio is defined for all n.
Mika Seppälä: Series
The Ratio Test30
Show that the series
xn
n!
n=0
∞
∑ converges for all values of x.
Solution
an+1
an
=
xn+1
n +1( )!
xn
n!
=n!
n +1( )!
xn+1
xn=
x
n +1n→ ∞ → 0 .
Hence
xn
n!n=0
∞
∑ converges regardless of the value of x.
Use the Ratio Test.