Worksheet on Section 11.2 - Answers
1. What is a sequence? What is an infinite series? What is the definition of the sum of an
infinite series? How does this relate to sequences?
A sequence is an infinite list of numbers. An
series is the sum of an infinite collection
Pinfinite
∞
of numbers.
P The definition of the sum of n=1 an is to define sn = a1 + a2 + · · · + an and
say s = ∞
n=1 = limn→∞ sn . The sequence of partial sums if a sequence, and the infinite sum
converges if and only if this sequence converges.
P The test for divergence states that if limn→∞ an
does not exist, or the limit is not zero, then ∞
n=1 an diverges.
(
1 if n is odd
2. (a) We have s1 = 1, s2 = 0, s3 = 1, s4 = 0, and so sn =
.
0 if n is even.
(b) Thus limn→∞ sn does not exist.
(c) Hence the series diverges.
3. (a) For this series, a = 3/10 and r, the ratio of any two consecutive terms is 1/10. Thus,
3/10
a
= 1−1/10
= 3/10
= 39 = 1/3.
since |r| < 1, the series converges and the sum is 1−r
9/10
(b) We can write this as 27 · 10−3 + 27 · 10−6 + 27 · 10−9 + · · · . Thus, a = 27 · 10−3 and r = 10−3 .
THe series converges to
27 · 10−3
27
1
=
= .
−3
1 − 10
999
37
(c) This series has a =
15
2
and r = 3/2. Since |r| > 1, the series diverges.
(d) This series has a = −6/5 and r = −6/5. Since |r| > 1, the series diverges.
(e) In this series the next term is made from the previous by multiplying by e/π. Thus, r = e/π
and a = π. Since e ≈ 2.718 and π ≈ 3.14, we have that |e/π| < 1 so the series converges, and
it converges to
a
π
π2
=
=
.
1−r
1 − e/π
π−e
4. (a) Let sn be the nth partial sum. Then
sn = (cos(π/1) − cos(π/2)) + (cos(π/2) − cos(π/3)) + · · · + (cos(π/n) − cos(π/(n + 1)).
All the terms cancel except the first cos(π) and the last cos(π/(n + 1)) and so sn = cos(π) −
cos(π/(n + 1)). As n → ∞ we get cos(π) − cos(0) = −1 − 1 = −2.
1
A
B
(b) We want (x+1)(x+2)
= x+1
+ x+2
. Multiplying we get A(x + 2) + B(x + 1) = 1. Setting
x = −1 we get A(−1 + 2) = 1 and so A = 1. Setting x = −2 we get B(−2 + 1) = 1 and so
1
1
1
B = −1. Thus, (x+1)(x+2)
= x+1
− x+2
.
1
2
1
Using this, if sn = a1 + · · · + an , then we have cancellation and get 21 − n+2
. Thus, limn→∞ sn =
1/2.
(c) The series is ln((n + 1)/n) = ln(n + 1) − ln(n). Thus,
sn = (ln(2) − ln(1)) + (ln(3) − ln(2)) + · · · + (ln(n + 1) − ln(n)).
Everything cancels except the first − ln(1) and the last ln(n + 1) and so sn = ln(n + 1) − ln(1) =
ln(n + 1). This tends to infinity and so the series diverges.
(d) Some tricky algebra shows that
√
√ √
√
√
√ n+1+ n
n+1− n=
n+1− n · √
√
n+1+ n
(n + 1) − n
=√
√
n+1+ n
1
=√
√ .
n+1+ n
Hence,
√
√
√
√
√
√
sn = ( 2 − 1) + ( 3 − 2) + · · · + ( n + 1 − n).
√
√
√
Everything cancells except the − 1 and the n + 1 and so sn = n + 1 − 1. Since sn → ∞,
the series diverges.
5. (a) We will try the test for divergence. We have limn→∞ cos(1/n) = cos (limn→∞ 1/n) =
cos(0) = 1. Since this is not zero, the test for divergence says it diverges.
P
P∞ 10n/2
1
(b) We get ∞
n=1 n(n+1) +
n=1 3n . The first series is a convergent telescoping series that sums
√
√
to 1. The second is a geometric series with a = 10/3 and r = 10/3. Since r2 = 10/9 > 1,
we have that |r| > 1 and so the geometric series part diverges. Thus, the whole series diverges.
(c) We have that an = 0 if n is odd andPan = 1 if n is even. Thus, limn→∞ an does not exist
and so the test for divergence says that ∞
n=1 an does not exist either.
P∞ (−1)n
P∞ 1
(d) The series is
n=1 2n . The first is a geometric series with a = 1/2 and
n=1 2n +
r = 1/2 which converges to a/(1 − r) = 1/2/(1 − 1/2) = 1. The second is a geometric series
−1/2
with a = −1/2 and r = −1/2 which converges to 1−(−1/2)
= (−1/2)/(3/2) = −1/3. Thus, the
whole series converges to 1 − 1/3 = 2/3.
6. (a) False. If an = 1 for all n, then
limn→∞ an = 1 converges. Then sn = a1 +a2 +· · ·+an = n.
P∞
The sequence sn diverges and so n=1 an diverges.
3
(b) This is true. It is a version of the test for divergence. The proof is here: Let sn =
a1 + a2 + · · · + an . If the series is convergent, then limn→∞ sn = s. This means that
lim an = lim sn − sn−1
n→∞
n→∞
= lim sn − lim sn−1
n→∞
n→∞
= s − s = 0.
(c) This is false.
The series 4(c) above we found to be a divergent telescoping series. However,
an = ln 1 + n1 . We have
1
lim an = ln lim 1 +
= ln(1) = 0.
n→∞
n→∞
n
7. We have p2 = a2 = 1 − 1/4 = 3/4. We have p3 = (1 − 1/4)(1 − 1/9) = (3/4)(8/9) = 3(2/9) =
6/9 = 2/3. We have p4 = p3 (15/16) = (2/3)(15/16) = 2(5/16) = 10/16 = 5/8.
Let qn =
n+1
.
2n
We’ll show that pn = qn for all n. We have q2 = 3/4 and this is p2 .
If pn = qn , then
pn+1
= pn · 1 −
1
(n + 1)2
n2 + 2n
= pn ·
(n + 1)2
n2 + 2n
= qn ·
(n + 1)2
n + 1 n(n + 2)
=
·
2n
(n + 1)2
n(n + 1)(n + 2)
=
2n(n + 1)2
n+2
= qn+1 .
=
2(n + 1)
Thus, pn = qn for all n and so
n+1
= 1/2.
n→∞ 2n
lim pn = lim qn = lim
n→∞
n→∞