PRACTICE PROBLEMS: SET 9
MATH 101: PROF. DRAGOS GHIOCA
1. Problems
Problem 1. Determine with proof whether the following sequences are convergent
or divergent. If a sequence is convergent, find its limit.
(a)
{
( 2
)}∞
n +1
−1
tan
n
n=1
(b)
{
1 + sin(n)
n2
}∞
(c)
{ −n }∞
e
n=0
(d)
{
}∞
2
(−1)n
n=2
n=1
Problem 2. Determine whether the following series are convergent or divergent.
(a)
∞
∑
ln(n)
n2
n=2
(b)
∞
∑
(−1)n
n+1
n=0
(c)
( )
1
cos
n
n=1
∞
∑
Problem 3. Compute the sum of each of the following series:
(a)
∞
∑
6 · 2n
n=0
(b)
∞
∑
5n+1
(
1
ln 1 − 2
n
n=2
1
)
2
MATH 101: PROF. DRAGOS GHIOCA
2. Solutions
Problem 1.
(a) We easily compute that
n2 + 1
1
= lim n + diverges to +∞.
n→∞
n→∞
n
n
On the other hand,
π
lim tan−1 (x) = .
x→+∞
2
Therefore
( 2
)
n +1
π
lim tan−1
= .
n→∞
n
2
(b) Using that −1 ≤ sin(n) ≤ 1 we obtain that
lim
1 + sin(n)
2
≤ 2.
2
n
n
= 0, we conclude by the Squeeze Theorem that
0≤
Since limn→∞
2
n2
1 + sin(n)
= 0.
n2
(c) Clearly, −n → −∞ as n → ∞. On the other hand,
lim
n→∞
lim ex = 0,
x→−∞
which proves that
lim e−n = 0.
n→∞
(d) If n is even, then n2 = n · n is also even, while if n is odd, then n2 = n · n
is also odd (note that product of two odd numbers is odd, while product of
two even numbers is even). So,
2
(−1)n = 1 if n is even, while
2
(−1)n = −1 if n is odd.
Therefore, this sequence alternates between −1 and 1 and thus it is divergent.
Problem 2.
(a) We let f (x) = ln(x)
x2 and note that the general term of the series is f (n).
We easily check that f (x) > 0 and
· x2 − ln(x) · 2x
1 − 2 ln(x)
=
.
x4
x3
Now, if x ≥ 2, then f ′ (x) < 0 since
f ′ (x) =
1
x
1 − 2 ln(x) ≤ 1 − 2 ln(2) = 1 − ln(4) < 0,
for x ≥ 2. Therefore we may apply the Integral Test in order to determine
the convergence or divergence of the above series; so, we analyze
∫ ∞
∫ t
ln(x)
ln(x)
dx = lim
dx .
2
t→∞
x
x2
2
2
PRACTICE PROBLEMS: SET 9
3
We first compute the antiderivative:
∫
ln(x)
dx
x2
∫
u
=
du using the substitution x = eu
eu
∫
=
ue−u du
∫
= −ue−u − −e−u du using integration by parts
= −ue−u − e−u + C
ln(x) 1
= −
− +C.
x
x
So,
∫
lim
t→∞
2
t
ln(x)
dx
x2
)
(
ln(x) 1 t
−
|
= lim −
t→∞
x
x 2
ln(t) 1 ln(2) 1
= lim −
− +
+
t→∞
t
t
2
2
1 + ln(2)
=
,
2
since limt→∞ ln(t)
= 0. Note that the last limit is indeed equal to 0 using
t
L’Hôpital’s Rule (being a limit of the form ∞/∞):
1
ln(t)
= lim t = 0.
t→∞
t→∞ 1
t
∫∞
This proves the improper integral 2 ln(x)
x2 dx is convergent, which in turn
proves that the given series is convergent as well.
(b) This is an alternating series which converges by the Alternating Series Test.
Indeed,
1
1
1
<
and lim
= 0.
n→∞ n + 1
n+2
n+1
(c) We compute
(
)
( )
1
1
= cos lim
= cos(0) = 1 ̸= 0,
lim cos
n→∞ n
n→∞
n
lim
which proves that the given series diverges, using the Divergence Test for
series.
Problem 3.
(a) The series is geometric; the first two terms are (in order) 65 and 12
25 . So,
6
using the notation for geometric series, we have a = 5 and r = 25 < 1
(hence the series is convergent), which allows us to conclude that
∞
6
∑
6 · 2n
5
=
n+1
5
1
−
n=0
2
5
= 2.
4
MATH 101: PROF. DRAGOS GHIOCA
(b) This series is telescoping. Indeed,
(
)
( 2
)
(
)
1
n −1
(n + 1)(n − 1)
ln 1 − 2 = ln
= ln
= ln(n−1)+ln(n+1)−2 ln(n).
n
n2
n2
So, we compute the N -th partial sum
sN
N
∑
(
1
=
ln 1 − 2
n
n=2
=
N
∑
)
(ln(n − 1) + ln(n + 1) − 2 ln(n))
n=2
= ln(1) − ln(2) − ln(N ) + ln(N + 1).
Now,
(
)
1
lim − ln(N ) + ln(N + 1) = lim ln 1 +
= ln(1) = 0.
N →∞
N →∞
N
So, we conclude that limN →∞ sN = − ln(2), which proves that
)
(
∞
∑
1
ln 1 − 2 = − ln(2).
n
n=2