Math 312, Intro. to Real Analysis: Homework #5 Solutions

Stephen G. Simpson

Friday, March 20, 2009

The assignment consists of Exercises 14.3, 14.4, 14.6, 14.13, 15.3, 15.4, 15.7 in the Ross textbook. Each problem counts 10 points.

In solving some of these exercises, we use the fact that ∑ 1/n^{s} is convergent if s > 1 and divergent if s ≤ 1. This result appears as Theorem 15.1 in the textbook.

14.3. (a) ∑ \frac{1}{\sqrt{n!}} converges by the ratio test: \frac{a_{n+1}}{a_{n}} = \frac{1}{\sqrt{n + 1}} → 0.

(b) ∑ \frac{2 + cos n}{3^{n}} converges by comparision with the geometric series

∑ \frac{3}{3^{n}} = 1 + \frac{1}{3} + \frac{1}{9} + ⋯ = \frac{3}{2}, because 1 ≤ 2 + cos x ≤ 3.

(c) ∑ 1/(2^{n} + n) converges by comparison with the geometric series

∑ 1/2^{n}, because 2^{n} + n ≥ 2^{n} for all n.

(d) ∑ (\frac{1}{2})^{n} (50 + \frac{2}{n}) converges by comparison with the geometric series

∑ 52 (\frac{1}{2})^{n}, because 50 + \frac{2}{n} ≤ 52.

(e) ∑ sin(nπ/9) diverges by Corollary 14.5, because lim sin(nπ/9) does not exist.

(f) ∑ \frac{100^{n}}{n!} converges by the ratio test: \frac{a_{n+1}}{a_{n}} = \frac{100}{n + 1} → 0.

14.4. (a) ∑ 1/(n+(−1)^{n})^{2} converges by comparison with the convergent series

∑ 1/(n − 1)^{2}, because (n + (−1)^{n})^{2} ≥ (n − 1)^{2} for all n.

(b) ∑ (\sqrt{n + 1} − \sqrt{n}) = ∑ 1/ (\sqrt{n + 1} + \sqrt{n}) diverges by comparison

with the divergent series ∑ 1/2\sqrt{n + 1}, because

\sqrt{n + 1} + \sqrt{n} ≤ 2\sqrt{n + 1} for all n.

(c) ∑ \frac{n!}{n^{n}} converges by the ratio test: \frac{a_{n+1}}{a_{n}} = (\frac{n}{n + 1})^{n} → \frac{1}{e} < 1.

14.6. Assume that ∑ |a_{n}| converges and that (b_{n}) is a bounded sequence. Let M be such that |b_{n}| ≤ M for all n. Then |a_{n}b_{n}| = |a_{n}||b_{n}| ≤ M|a_{n}|.

Hence ∑ |a_{n}b_{n}| is convergent, by comparison with the convergent series ∑ M|a_{n}| = M ∑ |a_{n}|. Thus ∑ a_{n}b_{n} is absolutely convergent. It follows by Corollary 14.7 that ∑ a_{n}b_{n} is convergent.

14.13. (a) ∑_{n=1}^{∞} (\frac{2}{3})^{n} = ∑_{n=0}^{∞} ar^{n} where a = \frac{2}{3} and r = \frac{2}{3}. By equation (2) on

page 91, the sum of this geometric series is ∑_{n=1}^{∞} (\frac{2}{3})^{n} = \frac{\frac{2}{3}}{1 − \frac{2}{3}} = 2.

Similarly with a = r = −\frac{2}{3} we have ∑_{n=1}^{∞} (−\frac{2}{3})^{n} = \frac{−\frac{2}{3}}{1 − (−\frac{2}{3})} = −\frac{2}{5}.

(b) By cancellation, the nth partial sum is

∑_{k=1}^{n} \frac{1}{k(k+1)} = ∑_{k=1}^{n} (\frac{1}{k} − \frac{1}{k+1})
= (1 − \frac{1}{2}) + (\frac{1}{2} − \frac{1}{3}) + ⋯ + (\frac{1}{n−1} − \frac{1}{n}) + (\frac{1}{n} − \frac{1}{n+1})
= 1 − \frac{1}{n+1}

for all n. Taking the limit as n → ∞ we obtain ∑_{k=1}^{∞} \frac{1}{k(k+1)} = 1.

(c) By cancellation, the nth partial sum is

∑_{k=1}^{n} \frac{k−1}{2^{k+1}} = ∑_{k=1}^{n} (\frac{k}{2^{k}} − \frac{k+1}{2^{k+1}} )
= (\frac{1}{2} − \frac{2}{4}) + (\frac{2}{4} − \frac{3}{8}) + ⋯ + (\frac{n−1}{2^{n−1}} − \frac{n}{2^{n}} ) + (\frac{n}{2^{n}} − \frac{n+1}{2^{n+1}} )
= \frac{1}{2} − \frac{n+1}{2^{n+1}}

for all n. Taking the limit as n → ∞ we obtain ∑_{k=1}^{∞} \frac{k−1}{2^{k+1}} = \frac{1}{2}.

(d) Using the result of part (c) we have

∑_{n=1}^{∞} \frac{n}{2^{n}} = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + ⋯
= 4 · (\frac{1}{8} + \frac{2}{16} + \frac{3}{32} + ⋯)
= 4 · ∑_{n=1}^{∞} \frac{n−1}{2^{n+1}}
= 4 · \frac{1}{2} = 2.

15.3. Letting u = log x we have du = dx/x. For s ≠ 1 we have

∫_{2}^{∞} \frac{dx}{x(log x)^{s}} = ∫_{log 2}^{∞} \frac{du}{u^{s}} = \frac{u^{−s+1}}{−s + 1}|_{log 2}^{∞}

and this is finite if s > 1, infinite if s < 1. Thus by the integral test we

see that ∑_{n=2}^{∞} \frac{1}{n(log n)^{s}} converges if s > 1 and diverges if s < 1. For the

special case s = 1 we have

∫_{2}^{∞} \frac{dx}{x log x} = ∫_{log 2}^{∞} \frac{du}{u} = log u|_{log 2}^{∞}

and this is infinite, hence ∑_{n=2}^{∞} \frac{1}{n log n} diverges by the integral test.

15.4. (a) ∑ 1/\sqrt{n} log n diverges by comparison with the divergent series ∑ 1/n.

This is because log n < \sqrt{n}, hence \sqrt{n} log n < n.

(b) ∑ log n/n diverges by comparison with the divergent series ∑ 1/n.

This is because log n/n > 1/n for all n.

(c) Letting u = log log x we have du = dx/x log x, hence

∫_{4}^{∞} \frac{dx}{x log x log log x} = ∫_{log log 4}^{∞} \frac{du}{u} = log u|_{log log 4}^{∞} = ∞

hence ∑ 1/n log n log log n diverges by the integral test.

(d) ∑ log n/n^{2} converges by comparison with the convergent series ∑ 1/n^{3/2}.

This is because log n < \sqrt{n}, hence log n/n^{2} < 1/n\sqrt{n} = 1/n^{3/2}.

15.7. Assume that (a_{n}) is a nonincreasing sequence of real numbers and ∑ a_{n} converges. We are going to prove that lim na_{n} = 0.

Note first that a_{n} ≥ 0 for all n. This is because, if a_{n} < 0 for some n, then since our sequence is nonincreasing we would have a_{m} ≤ a_{n} < 0 for all m > n, hence lim a_{m} ≠ 0 which would contradict Corollary 14.5.

Now, given ϵ > 0, apply the Cauchy criterion to find an N so large that a_{m} + a_{m+1} + ⋯ + a_{n} < ϵ whenever N < m < n. In particular we have

a_{N+1} + a_{N+2} + ⋯ + a_{n} < ϵ

and

a_{N+1} ≥ a_{N+2} ≥ ⋯ ≥ a_{n}

and

\frac{n}{2} < n − N

whenever n > 2N. From these observations it follows that

\frac{na_{n}}{2} < (n − N)a_{n} < ϵ

whenever n > 2N. This shows that lim na_{n}/2 = 0, and it is then immediate that lim na_{n} = 0, Q.E.D.