MA1S12: SOLUTIONS TO TUTORIAL 8
1. Determine if the following series are convergent.
∞
X
1
√
k
k=1
∞
X
5
3k
k=0
Solution: The first series is a geometric series since it has the form
∞
X
ark
k=0
with a = 5 and r = 31 . Since r < 1 it is convergent.
The second series is a p-series since it has the form
∞
X
1
kp
k=1
with p = 21 . Since p ≤ 1 the series is divergent.
2. Determine if the following series are convergent.
∞
X
e
∞
X
1
k
k=1
k=0
3k
5
+4
Solution: For the first series we apply the divergence test: if
lim uk 6= 0
k→∞
then the series
P∞
k=1
uk is divergent. In this case we have
1
lim e k = e0 = 1 6= 0
k→∞
Thus the series
P∞
k=1
1
e k is divergent.
For the second series we will use the comparison test: if 0 ≤ uk ≤
P
vk for all k and the larger series
vk is convergent then the smaller
1
2
MA1S12: SOLUTIONS TO TUTORIAL 8
series
P
uk is also convergent. In this case we have
0≤
5
5
≤
3k + 4
3k
P
5
We saw in Question 1 that the larger series ∞
k=0 3k is a convergent
P
5
geometric series. Thus the smaller series ∞
k=0 3k +4 is also convergent.
3. Determine if the following series are convergent.
∞
X
k=1
∞
X
1
k!
k=1
k2 − 3
k3 + k + 1
Solution: For the first series we will apply the limit comparison test.
When k is large the terms with the highest power will dominate the
2
−3
numerator and denominator so k3k+k+1
behaves like
P∞ 1
compare our series with k=1 k . Compute
ρ =
k2
k3
= k1 . We will
uk
k→∞ vk
lim
k2 − 3 k
k→∞ k 3 + k + 1 1
k 3 − 3k
= lim 3
k→∞ k + k + 1
1 − k32
= lim
k→∞ 1 + 12 + 13
k
k
=
lim
= 1
P
k2 −3
Since 0 < ρ < ∞ the limit comparison test says that ∞
k=1 k3 +k+1
P∞ 1
P
1
behaves the same way as ∞
k=1 k . We know
k=1 k is a divergent
P
k2 −3
p-series (with p = 1). Thus ∞
k=1 k3 +k+1 is also divergent.
MA1S12: SOLUTIONS TO TUTORIAL 8
3
For the second series we will apply the ratio test. Compute
uk+1
k→∞ uk
k!
= lim
k→∞ (k + 1)!
1
= lim
k→∞ k + 1
ρ =
lim
= 0
Since ρ < 1 the ratio test says that
P∞
1
k=1 k!
is convergent.
4. Determine if the following series are convergent.
k
∞ X
2k
7k + 3
k=1
∞
X
(−1)k
k=0
3k
Solution: For the first series we apply the root test. Compute
ρ = lim
k→∞
√
k
2k
2
= lim
k→∞ 7k + 3
k→∞ 7 +
uk = lim
Since ρ < 1 the root test says that
P∞
k=1
k
2k
7k+3
3
k
=
2
7
is convergent.
The second series contains negative terms so we will apply the
absolute convergence test. Note that
∞ ∞
X
(−1)k X
1
=
3k 3k
k=0
k=0
is a convergent geometric series with r = 31 < 1. This shows that
P∞ (−1)k
is absolutely convergent and hence also convergent.
k=0 3k