Math 312, Intro. to Real Analysis: Homework #3 Solutions

Stephen G. Simpson

Monday, February 16, 2009

The assignment consists of Exercises 7.3, 7.5, 8.2(b)(e), 8.3, 8.8, 9.3, 9.4, 9.12, 9.18 in the Ross textbook. Each problem counts 10 points.

7.3. (a) 1, (b) 1, (c) 0, (d) 1, (e) does not converge, (f) 1, (g) diverges to +∞, (h) does not converge, (i) 0, (j) 7/2, (k) diverges to +∞, (l) does not converge, (m) 0, (n) does not converge, (o) 0, (p) 2, (q) 0, (r) 1, (s) 4/3,

(t) 0.

7.5. (a) lim \sqrt{n^{2} + 1} − n = lim \frac{1}{\sqrt{n^{2} + 1} + n} = 0.

(b) lim \sqrt{n^{2} + n} − n = lim \frac{n}{\sqrt{n^{2} + n} + n} = lim \frac{1}{\sqrt{1 + 1/n} + 1} = \frac{1}{2}.

(c) lim \sqrt{4n^{2} + n} − 2n = lim \frac{n}{\sqrt{4n^{2} + n} + 2n} = lim \frac{1}{\sqrt{4 + 1/n} + 2} = \frac{1}{4}.

8.2. (b) The limit is 7/3. Given ϵ > 0 we want

|\frac{7n − 19}{3n + 7} − \frac{7}{3}| < ϵ.  (1)

This is equivalent to

|\frac{−106}{9n + 21}| < ϵ,

i.e.,

\frac{106}{9n + 21} < ϵ,

i.e.,

n > \frac{106}{9ϵ} − \frac{7}{3}.

The above reasoning works in reverse, so we have shown that (1) holds for all n > N where N = 106/9ϵ − 7/3. This proves the limit.

(e) The limit is 0. Given ϵ > 0 we want

|\frac{sin n}{n}| < ϵ.  (2)

Since | sin x| ≤ 1 for all x, (2) will hold provided 1/n < ϵ, i.e., n > 1/ϵ. Thus we may take N = 1/ϵ and this proves the limit.

8.3. Assume lim s_{n} = 0 and s_{n} ≥ 0 for all n. Given ϵ > 0, let N be so large that s_{n} < ϵ^{2} for all n > N. Then \sqrt{s_{n}} < ϵ for all n > N. This proves that lim \sqrt{s_{n}} = 0.

8.8. (a) We want |\sqrt{n^{2} + 1} − n| < ϵ. By algebra this is equivalent to

\frac{1}{\sqrt{n^{2} + 1} + n} < ϵ.  (3)

But clearly \sqrt{n^{2} + 1}+n > 2n, hence (3) will hold provided 1/2n < ϵ, i.e., n > 1/2ϵ. This proves the limit.

(b) We want |\sqrt{n^{2} + n} − n − 1/2| < ϵ. By algebra this is equivalent to

|\frac{n − \sqrt{n^{2} + n}}{2\sqrt{n^{2} + n} + 2n}| < ϵ.  (4)

But clearly n − \sqrt{n^{2} + n} < 0 and 2\sqrt{n^{2} + n} + 2n > 4n, so (4) will hold provided

\frac{\sqrt{n^{2} + n} − n}{4n} < ϵ,

i.e.,

\frac{\sqrt{1 + 1/n} − 1}{4} < ϵ,

i.e.,

n > \frac{1}{(1 + 4ϵ)^{2} − 1}.

This proves the limit.

(c) We want |\sqrt{4n^{2} + n} − n − 1/4| < ϵ. By algebra this is equivalent to

|\frac{2n − \sqrt{4n^{2} + n}}{4\sqrt{4n^{2} + n} + 8n}| < ϵ.  (5)

But clearly 2n − \sqrt{4n^{2} + n} < 0 and 4\sqrt{4n^{2} + n} + 8n > 16n, so (5) will hold provided

\frac{\sqrt{4n^{2} + n} − 2n}{16n} < ϵ,

i.e.,

\frac{\sqrt{1 + 1/4n} − 1}{8} < ϵ,

i.e.,

n > \frac{1}{4(1 + 8ϵ)^{2} − 4}.

This proves the limit.

9.3. Assume lim a_{n} = a and lim b_{n} = b. Then, using limit theorems,

lim \frac{a_{n}^{3} + 4a_{n}}{b_{n}^{2} + 1} = \frac{lim(a_{n}^{3} + 4a_{n})}{lim(b_{n}^{2} + 1)} = \frac{lim a_{n}^{3} + lim 4a_{n}}{lim b_{n}^{2} + 1} = \frac{a^{3} + 4a}{b^{2} + 1}.

9.4. Define a sequence (s_{n}) inductively by letting s_{1} = 1 and s_{n+1} = \sqrt{s_{n} + 1} for all n.

(a) The first four terms are s_{1} = 1, s_{2} = \sqrt{2}, s_{3} = \sqrt{\sqrt{2} + 1},

s_{4} = \sqrt{\sqrt{\sqrt{2} + 1} + 1}.

(b) Assume that lim s_{n} = α. By induction on n we can easily show that s_{n} ≥ 1 for all n, hence α ≥ 1. Also, we have lim s_{n+1} = α because (s_{n+1}) is essentially the same sequence as (s_{n}). Taking the limit of both sides of the equation s_{n+1} = \sqrt{s_{n} + 1} and using limit laws, we see that α = \sqrt{α + 1}. Hence α^{2} − α − 1 = 0, i.e., α is a solution of the equation x^{2} − x − 1 = 0. The solutions of this equation are

x = \frac{1 ± \sqrt{5}}{2}

but the solution with 1 − \sqrt{5} is ruled out because it is < 0. We conclude that

α = \frac{1 + \sqrt{5}}{2}.

9.12. Assume that L = lim |s_{n+1}/s_{n}| exists and that s_{n} ≠ 0 for all n.

(a) If L < 1, let a be such that L < a < 1. Then for all sufficiently large n we have |s_{n+1}/s_{n}| < a. Let N be so large that this holds for all n ≥ N. Then for all n > N we have

s_{n} = s_{N} · \frac{s_{N+1}}{s_{N}} · \frac{s_{N+2}}{s_{N+1}} · · · · · \frac{s_{n}}{s_{n−1}} ,

hence

|s_{n}| = |s_{N}| · |\frac{s_{N+1}}{s_{N}} | · |\frac{s_{N+2}}{s_{N+1}} | · · · · · |\frac{s_{n}}{s_{n−1}} | ,

hence

|s_{n}| ≤ |s_{N}| · a · a · · · · · a = |s_{N}| · a^{n−N} = \frac{|s_{N}|}{a^{N}} · a^{n}.

Since lim a^{n} = 0, it follows that lim s_{n} = 0.

(b) If L > 1, let t_{n} = 1/|s_{n}|. We have |t_{n+1}/t_{n}| = 1/|s_{n+1}/s_{n}|, hence lim |t_{n+1}/t_{n}| = 1/L < 1. Applying part (a) to the sequence (t_{n}) we get lim t_{n} = 0. It then follows by Theorem 9.10 that lim |s_{n}| = +∞.

9.18. (a) For a ≠ 1 we have

(1 − a)(1 + a + a^{2} + ⋯ + a^{n})
= (1 + a + a^{2} + ⋯ + a^{n}) − (a + a^{2} + a^{3} + ⋯ + a^{n+1})
= 1 − a^{n+1}

hence

1 + a + a^{2} + ⋯ + a^{n} = \frac{1 − a^{n+1}}{1 − a} .

(b) For |a| < 1 we have lim a^{n+1} = 0 in view of Basic Example 9.7(b). Therefore, from part (a) plus limit laws, we have

lim(1 + a + a^{2} + ⋯ + a^{n}) = \frac{1}{1 − a}.

(c) Letting a = 1/3 in part (b), we obtain

lim (1 + \frac{1}{3} + \frac{1}{9} + ⋯ + \frac{1}{3^{n}} ) = \frac{1}{1 − 1/3} = \frac{3}{2}.

(d) For a ≥ 1 we have 1 + a + a^{2} + ⋯ + a^{n} ≥ 1 + n for all n, hence

lim(1 + a + a^{2} + ⋯ + a^{n}) = +∞.