# Harrison Chapman

## Math 317: Homework 5

Due: Friday, October 11, 2019

Exercises:

1. Show $\limsup(s_n+t_n) \le \limsup s_n + \limsup t_n$ for bounded sequences $(s_n)$ and $(t_n)$. Hint: First show

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, so that

and applying the definition of $\limsup$ yields,

as claimed. $\Box$

2. 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

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

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

1. 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. 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. 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. Suppose $$\sum a_n = A$$ and $$\sum b_n = B$$ where $$A, B \in \mathbb R$$.

1. Prove that $$\sum(a_n+b_n) = A+B$$.

2. Prove that $$\sum ka_n = kA$$ for any $$k \in \mathbb R$$.

3. Is it the case that $$\sum a_nb_n = AB$$? Explain your reasoning.

1. Let $$\alpha_N = \sum_{n=1}^{N} a_n$$ and $$\beta_N = \sum_{n=1}^{N} b_n$$. Then

$\lim_{N\to\infty} \sum_{n=1}^N {a_n + b_n} = \lim_{N\to\infty}{\alpha_N + \beta_N} = \lim_{N\to\infty}{\alpha_N} + \lim_{N\to\infty}{\beta_N} = A+B$

2. Let $$\alpha_N = \sum_{n=1}^{N} a_n$$. Then

$\lim_{N\to\infty} \sum_{n=1}^N {k a_n} = \lim_{N\to\infty}{k \alpha_N} = k\lim_{N\to\infty}{\alpha_N} = kA$

3. No. Consider $$(a_n) = (1, 3, 0, \cdots)$$ and $$(b_n) = (1, 2, 0, \cdots)$$. Then $$A = 4$$ and $$B = 3$$ but $$\sum a_nb_n = 1+6 = 7 \ne 12$$.

4. For each of the following, determine whether the given series converges and justify your answer.

1. Converges by root test; $$\lim \vert a_n \vert^{1/n} = 1/2 < 1$$

2. Converges by ratio test; $$\lim \left\vert \frac{a_{n+1}}{a_n} \right\vert = \lim \left(\frac{n}{n+1}\right)^n = 1/e < 1$$

3. Converges by comparison to $$\sum 1/3^n$$, which is a convergent geometric series.