Math 115 HW #3 Solutions
From §12.2
20. Determine whether the geometric series
∞
X
en
3n−1
n=1
is convergent or divergent. If it is convergent, find its sum.
Answer: I can re-write the terms as
en
3n−1
Therefore, the series
=e
e n−1
en−1
=
e
.
3n−1
3
∞ ∞ ∞
X
X
X
e n−1
e n
en
=
e
=
e
,
n−1
3
3
3
n=1
n=1
n=0
where the second equality comes from shifting the index by one. Since
the geometric series
∞ X
e n
1
3
.
=
e =
3
1− 3
3−e
e
3
< 1, we know that
n=0
Therefore, the given series converges and the sum is given by
∞
∞ X
X
e n
en
3
3e
=
e
=e
=
.
n−1
3
3
3−e
3−e
n=1
24. Determine whether the series
n=0
∞
X
k(k + 2)
k=1
(k + 3)2
is convergent or divergent. If it is convergent, find its sum.
Answer: This series diverges. To see this, I will show that the terms in the sequence do not
go to zero:
k(k + 2)
k 2 + 2k
lim
=
lim
.
k→∞ (k + 3)2
k→∞ k 2 + 6k + 9
Dividing numerator and denominator by k 2 yields
1
1 + k2
1 + k2
k2
lim
= lim
k→∞ 12 1 + 6 + 92
k→∞ 1 + 6 +
k
k
k
k
9
k2
= 1.
Therefore, using the nth term test (a.k.a. Test for Divergence), the series diverges.
1
38. Determine whether the series
∞
X
ln
n=1
n
n+1
is convergent or divergent by expressing sn as a telescoping sum. If it is convergent, find its
sum.
Answer: We can re-write the terms in the series as
n
ln
= ln(n) − ln(n + 1).
n+1
Therefore, the partial sum
sn = a1 + a2 + a3 + . . . + an
= (ln(1) − ln(2)) + (ln(2) − ln(3)) + (ln(3) − ln(4)) . . . + (ln(n) − ln(n + 1))
= ln(1) + (− ln(2) + ln(2)) + (− ln(3) + ln(3)) + . . . + (− ln(n) + ln(n)) − ln(n + 1)
= ln(1) − ln(n + 1)
= − ln(n + 1).
Therefore,
∞
X
n=1
ln
n
n+1
= lim sn = lim (− ln(n + 1)) = −∞,
n→∞
n→∞
so the given series diverges.
48. Find the values of x for which the series
∞
X
(x − 4)n
n=1
converges. Find the sum of the series for those values of x.
Answer: Notice that this is a geometric series, so the series converges when |x − 4| < 1,
meaning that
3 < x < 5.
Moreover, for such values of x, the series converges to
1
1
=
.
1 − (x − 4)
4−x
70. If
P
an and
P
bn are both divergent, is
P
(an + bn ) necessarily divergent?
Answer: No. Let an = 1 for all n and let bn = −1. Then
∞
X
an =
n=1
and
∞
X
n=1
bn =
∞
X
1 diverges
n=1
∞
X
n=1
2
(−1) diverges
However,
∞
X
(an + bn ) =
n=1
∞
X
(1 + (−1)) =
n=1
∞
X
0=0
n=1
certainly converges.
From §12.3
16. Determine whether the series
∞
X
n=1
n2
n3 + 1
is convergent or divergent.
2
Answer: If we let f (x) = x3x+1 , then the terms of the series and the function f satisfy the
hypotheses of the Integral Test, so the series will converge if and only if
Z ∞
Z ∞
x2
f (x)dx =
dx
x3 + 1
1
1
is finite.
Letting u = x3 + 1, we have that du = 3x2 dx, so I can re-write the above integral as
∞
Z
1 ∞ du
1
,
=
ln |u|
3 u=2 u
3
2
which diverges since ln(u) → ∞ as u → ∞. Therefore, the series diverges by the Integral
Test.
22. Determine whether the series
∞
X
n=2
1
n(ln n)2
is convergent or divergent.
Answer: If we let f (x) = x(ln1x)2 , then the terms of the series and the function f satisfy the
hypotheses of the Integral Test, so the series will converge if and only if
Z ∞
Z ∞
1
f (x)dx =
dx
x(ln x)2
2
2
is finite.
Letting u = ln x, we have that du = x1 dx, so I can re-write the above integral as
Z
∞
u=ln 2
du
= − u−1
u2
∞
=
u=ln 2
1
,
ln 2
which is finite. Therefore, the series converges by the Integral Test.
3
30. Find the values of p for which the series
∞
X
ln n
n=1
np
is convergent.
Answer: When p ≤ 0 the terms in the series do not go to zero, so the series will diverge.
When p > 0, the function f (x) = lnxpx and the series satisfy the hypotheses of the Integral
Test, so the series will converge if and only if
Z ∞
Z ∞
ln x
f (x)dx =
dx
(*)
xp
1
1
is finite.
When p = 1, I can write the above integral as
Z ∞
ln x
dx.
x
1
Letting u = ln x, we have du = x1 dx, so this is equal to
∞
u2
udu =
2
u=0
Z
∞
,
0
which is infinite.
When p 6= 1, I will use integration by parts to evaluate the integral in (*). Letting u = ln x
and dv = dx
xp , I have that
u = ln x
du =
1
dx
x
dv = x−p dx
v=
x1−p
,
1−p
so the anti-derivative is given by
Z
Z
ln x
x1−p
1
1
dx = (ln x)
−
dx
p
x
1−p 1−p
xp
x1−p
1 x1−p
= (ln x)
−
1−p 1−p1−p
(1 − p) ln x − 1
.
= x1−p
(1 − p)2
Hence,
Z
1
∞
ln x
(1 − p) ln x − 1
dx = x1−p
p
x
(1 − p)2
which is finite only when 1 − p < 0.
Therefore, the series converges when p > 1.
4
∞
,
1
34. Find the sum of the series
P∞
n=1 1/n
5
correct to three decimal places.
Answer: If we estimate the sum by the nth partial sum sn , then we know that the remainder
Rn is bounded by
Z ∞
Z ∞
1
1
dx ≤ Rn ≤
dx.
5
5
n+1 x
n x
This means that
Z
∞
Rn ≤
n
1
1 1
dx = − 4
5
x
4x
∞
=
n
1
,
4n4
so the estimate will be accurate to 3 decimal places when this expression is less than 0.001.
In other words, we want to know for what n is it true that
1
1
<
.
4
4n
1000
Solving for n, we get that
n>
√
4
250 ≈ 3.98.
So letting n = 4, we have that
s4 =
1
1
1
1
1
1
1
+ 5 + 5 + 5 =1+
+
+
≈ 1.036
5
1
2
3
4
32 243 1024
is an estimate of the sum of the series that is correct to three decimal places.
From §12.4
18. Determine whether the series
∞
X
n=1
1
2n + 3
converges or diverges.
Answer: Use the Limit Comparison Test to compare this series to
1
2n+3
1
n→∞
n
lim
Therefore, since
diverges.
P
1
n
P
1
n.
We see that
n
1
= .
n→∞ 2n + 3
2
= lim
diverges, the Limit Comparison Test tells us that the series
26. Determine whether the series
∞
X
√
3
n=1
P
1
2n+3
n+5
n7 + n2
converges or diverges.
Answer: Use the Limit Comparison Test to compare this series to
n+5
√
3 7
n +n2
lim
1
n→∞
n4/3
P
1
.
n4/3
(n + 5)n4/3
n7/3 + 5n4/3
√
√
=
lim
.
3
n→∞ 3 n7 + n2
n→∞
n7 + n2
= lim
5
We see that
also
Therefore, dividing both numerator and denominator by n7/3 , we see that this limit is equal
to
1
7/3 + 5n4/3
n
1 + n5
7/3
q
=
lim
lim n 1 √
= 1.
3
n→∞ 3
n→∞
n7 + n2
1 + 15
n7/3
n
Therefore, since
also converges.
P
1
n4/3
converges, the Limit Comparison Test tells us that the given series
32. Determine whether the series
∞
X
1
1+1/n
n
n=1
converges or diverges.
Answer: Use the Limit Comparison Test to compare this series to
1
n1+1/n
lim
1
n→∞
n
= lim
n→∞
n
n1+1/n
= lim
n→∞
P
1
n.
We see that
n
1
= lim √
=1
n
1/n
n→∞
n
n·n
P1
√
since limn→∞ n n = 1. Therefore, since
n diverges, the Limit Comparison test tells us that
P
1
also
diverges.
n1+1/n
6