Harrison Chapman

Math 317: Homework 4

Due: Friday, March 1, 2019
1. (12.4) Show $\limsup(s_n+t_n) \le \limsup s_n + \limsup t_n$ for bounded sequences $(s_n)$ and $(t_n)$. You may use the results of Exercise 9.9 from the book without first proving them. Hint: First show

and then apply Exercise 9.9(c).

Proof. Let $N > 0$ and consider $k > N$. As $s_k \in \{s_n:n>N\}$, $s_k \le \sup\{s_n:n>N\}$. Similarly, $t_k \le \sup\{t_n:n>N\}$. So,

so that $\sup\{s_n:n>N\} + \sup\{t_n:n>N\}$ is an upper bound for the set $\{s_n + t_n:n>N\}$. In particular, this means that

Taking limits as $N \to \infty$ of both sides preserves the inequality (this is the result of Exercise 9.9(c)), so that

and applying the definition of $\limsup$ yields,

as claimed. $\Box$

2. (12.12) This question is worth double points. Let $(s_n)$ be any sequence of non-negative real numbers. Let $(\sigma_n)$ be defined by,

1. Show

Hint: For the last inequality, show first that $M > N$ implies

Proof. We have trivially that $\liminf \sigma_n \le \limsup \sigma_n$, so we need only prove the first and last inequalities. First we will prove the inequality:

Consider $N > 0$ and any $M > N$. Consider any $k > M$. We will bound $\sigma_k$ from above:

where we have used that (1) $\sum_{i=1}^{N}{s_i} \ge 0$ as all $s_n$ are nonnegative, (2) $% $, and (3) that for all $% $ we have $s_i \le \sup\{s_n : N > N\}$. So,

As $\frac{k-N}k \le 1$. This upper bound holds for all $k > M$, so we may conclude that

As this relationship holds for all $M > N$, taking limits of either side as $M \to \infty$ (we may do this as increasing $M$ preserves the property that $M > N$) we obtain,

so that $\limsup{\sigma_n} \le \sup\{s_n : n > N\}$.

Taking limits then as $N \to \infty$ we get that

as desired.

We now show the first inequality, which uses similar techniques but is slightly different. We take $N > 0$, any $M > N$, and consider arbitrary $k > M$. Then we examine a lower bound for $\sigma_k$;

where we have used that (1) $\sum_{i=1}^{N}{s_i} \ge 0$ as all $s_n$ are nonnegative, and (2) that for all $% $ we have $s_i \ge \inf\{s_n : N > N\}$. Notice now that $\frac{k-N}{k} = 1 - \frac Nk \ge 1 - \frac NM$, so that

This lower bound holds for all $k > M$ so we have proved that

Similar to before, taking limits as $M \to \infty$ yields that

and taking limits as $N \to \infty$ yields finally that

as desired. $\Box$

2. Show that if $\lim_{n\to\infty} s_n$ exists, then $(\sigma_n)$ converges and $\lim_{n\to\infty} \sigma_n = \lim_{n\to\infty} s_n$

Proof. If $\lim s_n$ exists then $\lim s_n = \liminf s_n = \limsup s_n$ and so by (a) $\lim s_n = \liminf \sigma_n = \limsup \sigma_n = \lim \sigma_n$. $\Box$

3. Give an example of a sequence $(s_n)$ so that $(\sigma_n)$ converges but $(s_n)$ does not.

Here is one example:

Let $s_n = (-1)^n$. Then $(s_n)$ diverges, but what of $\sigma_n$? For even $n$ we can see that $\sigma_n = 0$ and odd $n$ we have that $\sigma_n = -\frac 1n$. This means that $\limsup \sigma_n = \lim 0 = 0$ and $\liminf \sigma_n = \lim \left(-\frac 1n\right) = 0$ and we conclude that $\lim \sigma_n = 0$.

3. (14.1–4) For each of the following, determine whether the given series converges and justify your answer.

1. Converges by the ratio test. We have the ratios $\frac{(n+1)^3}{3^{n+1}}\frac{3^n}{n^3} = \frac{(n+1)^3}{3n^3}$, whose $n \to \infty$ limit is $% $.

2. Converges by comparison, as $% $ whose corresponding series is a convergent geometric series.

3. Converges by the ratio test. We have the ratios

whose $n \to \infty$ limit is $% $.

4. (14.8) Show that if $\sum a_n$ and $\sum b_n$ are convergent series of nonnegative numbers, then $\sum \sqrt{a_n b_n}$ converges. Hint: Show $\sqrt{a_n b_n} \le a_n + b_n$ for all $n$.

Proof. We claim that $% $, from which convergence of the series follows from the comparison test.

Observe that as both $a_n$ and $b_n$ are nonnegative,

proving the claim. $\Box$