Section 11.5 Alternating Series
EX.11
Let bn = n2
n3+ 4. For n > 4, n3+ n
n3+ 4 > 1 ⇒ n2
n3+ 4 > 1
n + 1 = (n + 1)2
(n + 1)3 > (n + 1)2 (n + 1)3+ 4 Thus bn > bn+1 for all n > 4. Also, lim
n→∞bn = lim
n→∞
1/n
1 + 4/n3 = 0. Therefore the given series converges by the Alternating Series Test.
EX.14
Because lim
n→∞arctan n = π 2, lim
n→∞(−1)narctan n 6= 0 (the limit does not exist) and hence the given series diverges.
EX.20
Let bn =√
n + 1 −√
n. Then
bn = 1
√n + 1 +√
n > 1
√n + 2 +√
n + 1 = bn+1 Also, lim
n→∞bn = lim
n→∞
√ 1
n + 1 +√
n = 0. Therefore the given series converges by the Alternating Series Test.
EX.35
Let s2n =
2n
P
k=1
bk be the partial sum of the series. Then
s2n =
1 + 1
3+ · · · + 1 2n − 1
− 1 22 + 1
42 + · · · + 1 (2n)2
=
1 + 1
3+ · · · + 1 2n − 1
− 1 4
1 + 1
22 + · · · + 1 n2
>
1 + 1
3+ · · · + 1 2n − 1
− 1 4
∞
X
k=1
1 k2
=
n
X
k=1
1
2k − 1 − c 4
1
where c =
∞
P
k=1
1
k2 is finite since this p-series is convergent (p = 2 > 1). Now, by the Limit Comparison Test with the divergent p-series
∞
P
k=1
1
k (p = 1),
lim
k→∞
1/(2k − 1)
1/k = 1
2 > 0 ⇒
∞
X
k=1
1
2k − 1 = ∞
Therefore lim
n→∞s2n =P∞ k=1
1
2k − 1−c
4 = ∞. Also, since s2n+1 > s2n for all n ≥ 1, we have lim
n→∞s2n+1 =
∞. In conclusion,
∞
P
n=1
bn = ∞. the series diverges. The Alternating Series Test fails because bn is not decreasing (even we only consider its tail!). To see this, note that
n→∞lim b2n+1
b2n = lim
n→∞
b2n−1 b2n = ∞
⇒ ∃ an integer N such that b2n−1 > b2n and b2n+1 > b2n for all n > N
EX.36
(a)
s2n = 1 − 1 2 +1
3 −1
4 + · · · + 1
2n − 1 − 1 2n
=
1 + 1
2 +1 3 +1
4 + · · · + 1
2n − 1+ 1 2n
− 2 1 2+ 1
4+ · · · + 1 2n
=
1 + 1
2 +1 3 +1
4 + · · · + 1
2n − 1+ 1 2n
−
1 + 1
2 + · · · + 1 n
= h2n− hn
(b) Because the limits lim
n→∞hn− ln n and lim
n→∞h2n− ln(2n) exist,
n→∞lim s2n= lim
n→∞(h2n− hn)
= lim
n→∞[(h2n− ln(2n)) − (hn− ln n) + (ln(2n) − ln n)]
= lim
n→∞(h2n− ln(2n)) − lim
n→∞(hn− ln n) + lim
n→∞(ln(2n) − ln n)
= γ − γ + ln 2
= ln 2
2
