These are practice problems for series and power series. For all of the
following series check (and fully justify why) whether the series converges
or diverges. For problems with a star, if the series converges find its sum.
If the problem is about a power series find both its radius and interval of
convergence
Brief Solutions
1.
∞
X
2 ln n
n=1
n6
P∞
P
2n
converges by the Comparison Test, comparing with ∞
n=1
n=1 n6 = 2
To see why we can compare with the latter check that ln x < x, (hint:
do monotonicity analysis on f (x) = ln x − x )
2.
∞
X
cos2 (5n)
n=1
n3
converges by the Comparison Test, comparing with
−1 ≤ cos(5n) ≤ 1 ⇒ 0 ≤ cos2 (5n) ≤ 1
3.
1
n=1 n3
since
∞
X
(−4)n
n n
6
n=1
converges by the Ratio Test, the limit being
4.
P∞
4
6
=
2
3
< 1.
∞
X
7
n5n
n=1
converges byPthe Comparison
Test, since n57n ≤ 57n and the series
P
P
∞
∞
∞
1
1 n
7
n=1 5n = 7
n=1 5n = 7
n=1 ( 5 ) which is a converging geometric
series. Alternatively, one can use the Ratio Test.
5.
∞
X
n=1
4+
1
√
4
n3
diverges by the Limit Comparison Test; notice that for large n the
fraction behaves like n31/4 whose corresponding series diverges (p-series
with p < 1). Since we cannot use the comparison test (check that the
1
1
.
n5
inequality is in the unhelpful direction), the problem calls for the limit
comparison test. Check that limn
1
√
4
4+
n3
1
√
4
= 1 > 0 hence by the limit
n3
comparison test our series also diverges.
6.
∞
X
(−1)n
n=1
n3 2n
n!
converges by the Ratio Test (limit is 0)
7.
converges since
series
P∞
8n
n=1 32n
=
8.
∞
X
8n
32n
n=1
P∞ 8n
n=1 9n
∞
X
n=1
which is a convergent geometric
2n2
9n2 − 7
diverges by the Divergence Test, since an =
9.
2n2
9n2 −7
→
2
9
6= 0
∞
X
10 + 9n
n=1
5 + 8n
n
diverges by the Divergence Test, since an = 10+9
→ ∞ 6= 0 (which
5+8n
you can see either by using L’ Hospital’s rule or by diving both numerator and denominator by 9n )
10.
∞
X
6n
6n+4
n=1
diverges by the Divergence Test, since an =
11.
∞
X
9n + 2n
n=1
10n
6n
6n+4
=
1
64
→
1
64
6= 0
[∗]
P
P
converges since we know that if both P an and
P
P bn converge then
is equal to
an +
bn . Here, we have
n + bn ) converges
P∞ 9 and
P
P(a
∞ 9n +2n
∞
2 n
n
=
(
)
+
(
)
and
the
two
series
are converging
n=1 10
n=1 10
n=1 10n
2
a
geometric ones. For the sum, use the geometric sum formula = 1−r
9 1
where a, r = 10 , 5 respectively (check why; hint: summation starts
from 1, not 0)
12.
∞
X
(−1)n
n=1
n3 + 1
n3 − 7
3
diverges by the Divergence Test, since for even n an = nn3 +1
→ 1,
−7
n3 +1
while for odd n an = (−1) n3 −7 → −1 so an doesn’t have a limit.
13.
∞
X
n!(x − 4)n
n=1
Use the Ratio Test; observe than unless x = 4, the limit is ∞. So the
power series converges only for x = 4
14.
∞
X
(−1)n n7n xn
n=1
1
7
Use the Ratio Test; R = but none of the endpoints gives a convergent
numerical series (same idea as the next solution) so the interval of
convergence is (− 71 , 17 )
15.
∞
X
xn (n + 9)
(−1)n
5n
n=1
Use the Ratio Test; Radius of convergence is R = 5 but neither of
the endpoints ±5 gives a convergent series (use the divergence criterion again taking cases for even and odd terms). So, the interval of
convergence is (−5, 5).
16.
∞
X
(−1)n
n=1
cos n
n5
converges; we will show that this series is absolutely convergent, thus
is
Consider the series of the absolute values which becomes
Pconvergent.
∞ | cos n|
.
Now,
by the comparison test (can now use since my series
n=1 n5
has non-negative terms) since 0 ≤ | cos n| ≤ 1 the series of the absolute
values converges by comparison to a converging p-series (p = 5). We
have shown that the series converges absolutely and hence converges.
3