MATH 101 V2A
March 11th – Practice problems
Hints and Solutions
1. Determine the convergence of each of the following series.
(a)
∞
X
(3n + 1)n
.
n3n
n=1
Solution: Since
s
n
(3n + 1)n
n3
n
=
3n + 1
n3
and
3n + 1
= 0 < 1,
n3
it follows from the Root Test that the series converges.
lim
n→∞
(b)
∞
X
1
.
0.5n − 1
n=3
∞
1 X1
diverges (harmonic series), it follows
0.5 n=3 n
from the Comparison Test that the series diverges.
Solution: Since
(c)
1
0.5n−1
>
1
0.5n
for all n ≥ 3, and
∞
X
sin2 (n)
.
2n
n=1
∞
X
1
is a convergent geometric series, it follows
n
2
n=1
from the Comparison Test that the series converges.
Solution: Since,
sin2 (n)
2n
≤
1
2n
for all n ≥ 1, and
∞
X
n2 n!
(d)
.
2n!
n=1
Solution: Since
(n+1)2 (n+1)!
2(n+1)!
n2 n!
2n!
=
(n + 1)2 (n + 1)!2n!
(n + 1)3
=
,
n2 2n!
n2 n!2(n+1)!
and
(n + 1)3
= 0 < 1,
n→∞ n2 2n!
it follows from the Ratio Test that the series converges.
lim
p
P∞
2. Prove the Root Test: Suppose
limn→∞ n |an | = L. Then, if L < 1, the series n=1 |an | converges,
P∞
and if L > 1, the series n=1 |an | diverges.
Solution: Suppose L < 1 and let > 0 be sufficiently small so that 0 < L + < 1 (for instance, take
= 1−L
2 ). Then, there is a number N such that, for all n ≥ N ,
0<L−<
p
n
|an | < L + ,
so
(L − )n < |an | < (L + )n .
Hence,
∞
X
|an | ≤
n=N
∞
X
(L + )n ,
n=N
which is a convergent geometric series. It follows from the Comparison Test that the series
∞
X
|an |
n=1
converges.
Now suppose L > 1 and let > 0 be such that L − > 1 (for instance, take = L−1
2 ). Then, following
the same argument as above we get that there is a number N such that, for all n ≥ N ,
(L − )n < |an | < (L + )n .
So
∞
X
|an | >
n=N
∞
X
(L − )n ,
n=N
which is a divergent geometric series. So, it follows from the Comparison Test that the series
∞
X
|an |
n=1
diverges.
3. Using partial sums, explain why the series
P∞
n
n=1 (−1)
Solution: Let sN denote the N th partial sum. Then
(
N
X
−1
n
sN =
(−1) =
0
n=1
does not converge absolutely OR conditionally.
if N is odd
if N is even.
Hence, {sN } = {−1, 0, −1, 0, −1, 0, . . .}, so clearly the sequence of partial sums does not converge (in
any way).
2