Strategy for Testing Series: Solutions
1. Since (−5)−n = (−1/5)n , this is a geometric series. Because |−1/5| < 1, it converges.
2. Since n2 < n2 + 6n = n(n + 6) for all n ≥ 0, we have
1
1
< 2.
n(n + 6)
n
P
Because
1/n2 converges (it’s a p-series with p = 2 > 1), the comparison test
P
implies that
1/(n(n + 6)) also converges.
3. Clearly, the sequence an = 1/(50n) is decreasing and converges to 0. Thus, the
P
alternating series test implies that (−1)n+1 an converges.
4. Using l’Hôpital’s rule, we have
√
√
1
√
n
n
2 n
lim
= lim 1 = lim
= ∞.
n→∞ ln n
n→∞
n→∞ 2
n
√
√
P
n
Hence, the limn→∞ (−1)n ln n doesn’t exist and therefore the series (−1)n+1 ln nn diverges.
5. Applying the ratio test, we have
r
r n+1
rnr
n
(n+1)r
= r < 1,
lim rn = lim
r = lim r
n→∞ (n + 1)
n→∞
n→∞
n+1
nr
and therefore the series converges.
6. If f (n) = (n2 − n)−1/2 , then f ′ (n) = − 21 (n2 − n)−3/2 (2n − 1). Thus, when n > 1,
f ′ (n) < 0 and f (n) is decreasing. Moreover, we have
1
= 0.
n→∞
n→∞
n(n − 1)
P p
Therefore, the alternating series test implies
1/ n(n − 1) converges.
3
3
7. Since n < n + 2, we have
√
√
n ln(n)
n ln(n)
<
.
3
n +2
n3
lim f (n) = lim p
1If
Furthermore, because ln(n) < n for all n > 0 1, we obtain
√
√
n ln(n)
nn
< 3 = n−3/2 .
3
n +2
n
P −3/2
Now, the series
n
converges (it’s a p-series
with p = 3/2 > 1) and hence the
P √n ln(n)
comparison test implies that the series
also converges.
n3 +2
x
f (x) = lnxx , then f ′ (x) = 1−ln
x2 . It follows that f (x) has a unique critcal point at x = e. Since
f ′ (x) > 0 when x < e and f ′ (x) < 0 when x > e, the first derivative test implies that f (x) has a global
maximum at e. Hence, lnxx = f (x) ≤ f (e) = lnee < 1 which yields ln x < x for all x > 0.
page 1 of 7
MAT V1102 – 004
Solutions: page 2 of 7
8. Since ex is a strictly increasing function, e1/n ≤ e for all n ≥ 1. Hence, we have
e1/n
e
≤ 3/2 .
3/2
n
n
P −3/2
Since
en
converges (it’s a p-series with p = 3/2 > 1), the comparison test
P 1/n −3/2
implies that
e n
also converges.
(n+2)(n+3)
9. If f (n) = (n+1)3 then
f ′ (n) =
(2n + 5)(n + 1)3 − 3(n2 + 5n + 6)(n + 1)2
n2 + 8n + 13
=
−
.
(n + 1)6
(n + 1)4
When n ≥ 0, f ′ (n) < 0 and f (n) is decreasing. Moreover,
(n + 1)(n + 2)
n2
=
lim
= 0.
n→∞
n→∞
n→∞ n3
(n + 1)3
P
Therefore, the alternating series test implies (−1)n (n+1)(n+2)
converges.
(n+1)3
10. Since
n n
n
n
nn n ···
> 1,
=
n!
n
n−1
n−2
2
1
P (−1)n nn
n n
diverges.
the limit limn→∞ (−1)n! n is not zero and hence the series
n!
11. For all n > 2, we have
n n − 1 n − 2 2 1 2 1 2
n!
···
<
= 2.
=
n
n
n
n
n
n
n
n
n
n
P −2
Since
2n converges (it’s a p-series with p = 2 > 1), the comparison test implies
P n!
that
converges. Finally, the absolute convergence test implies that the series
nn
P
n n!
(−1) nn also converges.
12. Applying the ratio test, we have
lim f (n) = lim
en+1
(n+1)!
lim en
n→∞
n!
e
= 0 < 1,
n→∞ n + 1
= lim
P en
and hence the series
converges.
n!
13. Applying the ratio test, we have
(n+1)2 (n+1)!
(2n+2)!
lim
n2 n!
n→∞
(2n)!
and therefore the series
n3
(n + 1)2 (n + 1)
=
lim
= 0 < 1,
n→∞ n4
n→∞ (2n + 2)(2n + 1)n2
= lim
P
n2 n!
(2n)!
converges.
Solutions: page 3 of 7
MAT V1102 – 004
14. Since n! < n! + 2, we have
3n
3n
<
.
n! + 2
n!
P 3n
Now, applying the ratio test to the series
, we obtain
n!
3n+1
(n+1)!
lim 3n
n→∞
n!
3
= 0 < 1, .
n→∞ n + 1
= lim
P 3n
P 3n
converges and the comparison test implies that
also
Thus, the series
n!
n!+2
converges.
15. Applying the ratio test, we have
n
n+1
n+1
ln n
(ln(n+1))n+1
= lim
·
.
lim
n
n→∞ n ln(n + 1)
n→∞
ln(n + 1)
(ln n)n
ln n
< 1. Hence,
Since ln x is a strictly increasing function, ln n < ln(n + 1) and ln(n+1)
n
n+1
ln n
n+1
1
0 ≤ lim
·
≤ lim
= lim
= 0.
n→∞ n ln(n + 1)
n→∞ n ln(n + 1)
n→∞ ln(n + 1) + n
ln(n + 1)
n+1
n
P n
ln n
n+1
· ln(n+1)
converges.
= 0 < 1 and the series
Therefore, limn→∞ n ln(n+1)
(ln n)n
16. Applying the ratio test, we have
(n+1)6 5n+1
(n+2)!
lim
n6 5n
n→∞
(n+1)!
5(n + 1)6
5n6
=
lim
= 0 < 1,
n→∞ n6 (n + 2)
n→∞ n7
= lim
P n6 5n
converges.
and hence the series
(n+1)!
17. Applying the ratio test, we have
en+1
(ln(n+1))n+1
lim
en
n→∞
(ln n)n
e
= lim
n→∞ ln(n + 1)
ln n
ln(n + 1)
n
.
ln n
< 1.
Because ln x is a strictly increasing function, ln n < ln(n + 1) and thus ln(n+1)
It follows that
n
e
ln n
e
0 ≤ lim
≤ lim
= 0.
n→∞ ln(n + 1)
n→∞ ln(n + 1)
ln n + 1
n
P en
e
ln n
Therefore, limn→∞ ln(n+1)
=
0
<
1
and
the
series
converges.
ln n+1
(ln n)n
18. Since | sin(nπ/7)| ≤ 1, we have
sin(nπ/7) ≤ 1 .
n3
3
n
MAT V1102 – 004
19.
20.
21.
22.
Solutions: page 4 of 7
P −3
Because the series
n converges
(it’s a p-series and p = 3 > 1), the comparison
P sin(nπ/7) test implies that the series
n3 converges. Finally, the absolute convergence
P sin(nπ/7)
converges.
test implies
n3
Applying the ratio test, we have
(−2)n+1 (n+1)! 2
= lim
lim = 0 < 1,
n
(−2) n→∞
n→∞ n + 1
n! P (−2)n
converges.
and therefore the series
n!
For all positive integers n, we have
n
1
1+
≥ 1.
n
n
n
P
6= 0 and hence the series
(−1)n 1 + n1
It follows that limn→∞ (−1)n 1 + n1
diverges.
Applying the ratio test, we have
(−1)n+1 (2n+3)! (2n + 1)!
1
= lim
lim =
lim
= 0 < 1.
n
(−1) n→∞
n→∞ (2n + 3)!
n→∞ (2n + 3)(2n + 2)
(2n+1)! P (−1)n
Therefore, the series
converges.
(2n+1)!
Using L’Hôpital’s rule, we have
1
ln n
= lim n1 = lim ln n = ∞ .
n→∞
n→∞ ln(ln n)
n→∞
n ln n
P ln n
diverges.
Therefore, the series
ln(ln n)
√
x is a strictly increasing function, the inequality n3 < n(n + 1)(n + 2) implies
23. Since p
3/2
n < n(n + 1)(n + 2) and
lim
0< p
1
< n−3/2 .
n(n + 1)(n + 2)
P −3/2
Because
n
converges (it’s a p-series with p = 3/2 > 1), the comparison test
P
1
√
implies that the series
also converges.
n(n+1)(n+2)
f (n) = 21 n2 − 1
is positive; f ′ (n) = n is positive and f (2) = 1.
24. For n ≥ 2, the function
Adding 12 n2 to both sides of the inequality 0 < 21 n2 − 1 yields 12 n2 < n2 − 1 for n ≥ 2.
√
Since the square root function is strictly increasing, we have √12 n < n2 − 1 and
1
1
√
<√
2
n n −1
2n2
Solutions: page 5 of 7
MAT V1102 – 004
P 1
√
for n ≥ 3. Since
converges (it’s a p-series with p = 2 > 1), the comparison
P 2n12
√
also converges.
test implies that
n n2 −1
25. Using L’Hôpital’s rule, we have
√
r
√3
3 n+1
n
2 n+1
= lim
= 3 6= 0 .
lim √
= 3 lim
1
√
n→∞
n→∞
n→∞ n + 1
n+1
2 n
√
√
P
n+1
n+1
Therefore, limn→∞ (−1)n+1 3√n+1
6= 0 which implies that the series (−1)n+1 3√n+1
diverges.
26. If f (n) = ln(1 + 1/n) then
1
1
1
′
.
f (n) =
− 2 =−
n
n2 + n
1 + n1
When n > 0, f ′ (n) < 0 and therefore f (n) is decreasing. Moreover,
1
lim ln 1 +
= ln 1 = 0 ,
n→∞
n
P
and hence the alternating series test implies (−1)n ln 1 + n1 converges.
27. If n is a positive integer then cos(nπ) = (−1)n . Clearly
1
1
1
<
and lim = 0 .
n→∞ n
n+1
n
P cos nπ
Therefore, the alternating series test implies that the series
converges.
n
28. For n ≥ 3, the function f (n) = 12 n3 − 5 is positive; f ′ (n) = 23 n2 is positive and
− 5 > 0. Adding 21 n3 to both sides of the inequality 0 < 21 n3 − 5 yields
f (3) = 27
2
1 3
n < n3 − 5 which implies
2
1
2
<
n3 − 5
n3
P −3
for n ≥ 3. Since
2n converges (it’s a p-series with p = 3 > 1), the comparison
P 1
also converges.
test implies that
n3 −5
2
−1
29. If f (n) = (n + 2n + 1) = (n + 1)−2 then f ′ (n) = −2(n + 1)−3 . When n > 0,
f ′ (n) < 0 and f (n) is decreasing. Moreover,
1
= 0,
n→∞ n2 + 2n + 1
P (−1)n−1
converges.
and therefore the alternating series test implies
n2 +2n+1
30. For all n ≥ 1, we have
(2n)!
2n − 1
n+1
(2n) · (2n − 1) · · · · · (n + 1)
2n
···
≥ 1.
=
=
2n · n! · n
2n n
2n
2
2
P
Hence limn→∞ (−1)n 2n(2n)!
=
6
0
and
the
series
(−1)n 2(2n)!
n ·n!·n diverges.
·n!·n
lim
MAT V1102 – 004
Solutions: page 6 of 7
31. Since 5n < 5n + n, we have
n
1
1
2n+1
2n+1
2
< n which implies
< n =2
.
n
n
n+5
5
n+5
5
5
P 2 n
Because 2/5 < 1 the geometric series
2 5 converges. Applying the comparison
P 2n+1
also converges. Finally, |(−2)n+1 | = 2n+1 so the
test, we conclude that
n+5n
P (−2)n+1
converges.
absolute convergence test implies that the series
n+5n
32. Since | sin n| < 1, we have
(−1)n sin n ≤ 1 .
n2
2
n
P −2
Since the series
n converges (it’s a p-series with p = 2 > 1), the comparison test
P sin n
also converges. Finally, the absolute convergence test
implies that the series
n2
P
n sin n
implies that (−1) n2 converges.
33. Since en < en + 1, we have
2
2
< n.
n
e +1
e
P −n
Since the series 2 e converges (it’s a geometric series and e−1 < 1), the comparP 2
also converges.
ison test implies that the series
en +1
34. Since
− n12 cos n1
sin n1
1
1
lim n sin
= lim
= lim cos
= lim
= cos 0 = 1 6= 0 ,
1
1
n→∞
n→∞
n→∞
n→∞
n
n
−
2
n
n
P
the series
n sin n1 diverges.
35. Because f (x) = x(ln1x)p is positive, continuous and decreasing on [2, ∞), we may apply
the integral test. Since p > 1, we have
b
Z ∞
1
1
1−p
dx = lim
(ln n)
b→∞ 1 − p
x(ln x)p
2
0
1
1
=
(ln 2)1−p + lim
(ln b)1−p
b→∞
p−1
1−p
1
=
,
(p − 1)(ln 2)p−1
P 1
converges.
and therefore the series
n(ln n)p
√
√
−1
36. If f (n) = ( n + n + 1) then
√
√
1
1
−2
′
√ + √
.
f (n) = −( n + n + 1)
2 n 2 n+1
Solutions: page 7 of 7
When n > 0, f ′ (n) < 0 and f (n) is decreasing. Moreover,
1
√
= 0,
lim √
n→∞
n+ n+1
P (−1)n
√ √
so the alternating series test implies
converges.
n+ n+1
MAT V1102 – 004