# 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

$\sup\{s_n+t_n : n > N\} \le \sup\{s_n:n>N\} + \sup\{t_n:n>N\}.$

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,

$s_k + t_k \le \sup\{s_n:n>N\} + \sup\{t_n:n>N\},$

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

$\sup\{s_n + t_n:n>N\} \le \sup\{s_n:n>N\} + \sup\{t_n:n>N\}.$

Taking limits as $$N \to \infty$$ of both sides preserves the inequality, so that

\begin{align*} \lim_{N\to\infty}\sup\{s_n + t_n:n>N\} &\le \lim_{N\to\infty}(\sup\{s_n:n>N\} + \sup\{t_n:n>N\}) \\ &= \lim_{N\to\infty}\sup\{s_n:n>N\} + \lim_{N\to\infty}\sup\{t_n:n>N\}, \end{align*}

and applying the definition of $$\limsup$$ yields,

$\limsup s_n \le \limsup s_n + \limsup t_n,$

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,

$\sigma_n = \frac{s_1 + s_2 + \cdots + s_n}{n}.$
1. Show

$\liminf s_n \le \liminf \sigma_n \le \limsup \sigma_n \le \limsup s_n$

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

$\sup \left\{\sigma_n : n > M\right\} \le \frac 1M \left(s_1+s_2+\cdots+s_N\right) + \sup \left\{ s_n : n > N \right\}$
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:

$\limsup \sigma_n \le \limsup s_n.$

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

\begin{align} \sigma_k &= \frac 1k \sum_{i=1}^{k}{s_i} \\ &= \frac 1k \sum_{i=1}^{N}{s_i} + \frac 1k \sum_{i={N+1}}^{k}{s_i} \\ &\le \frac 1M \sum_{i=1}^{N}{s_i} + \frac 1k \sum_{i={N+1}}^{k}{\sup\{s_n : n > N\}}, \end{align}

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

\begin{align} \sigma_k &\le \frac 1M \sum_{i=1}^{N}{s_i} + \frac {k-N}{k} {\sup\{s_n : n > N\}} \\ &\le \frac 1M \sum_{i=1}^{N}{s_i} + {\sup\{s_n : n > N\}}, \end{align}

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

$\sup\{\sigma_n : n > M\} \le \frac 1M \sum_{i=1}^{N}{s_i} + {\sup\{s_n : n > N\}}.$

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,

\begin{align} \lim_{M\to\infty}\sup\{\sigma_n : n > M\} &\le \lim_{M\to\infty}\left(\frac 1M \sum_{i=1}^{N}{s_i}\right) + \lim_{M\to\infty}{\sup\{s_n : n > N\}} \\ &\le 0 + \sup\{s_n : n > N\}, \\ \end{align}

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

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

\begin{align} \lim_{N \to \infty}\limsup \sigma_n &\le \lim_{N \to \infty}\sup\{s_n : n > N\} \\ \limsup \sigma_n &\le \limsup {s_n} \end{align}

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 $$\sigma_k$$;

\begin{align} \sigma_k &= \frac 1k \sum_{i=1}^{k}{s_i} \\ &= \frac 1k \sum_{i=1}^{N}{s_i} + \frac 1k \sum_{i={N+1}}^{k}{s_i} \\ &\ge 0 + \frac 1k \sum_{i={N+1}}^{k}{\inf\{s_n : n > N\}} \\ &= \frac{k-N}{k}\inf\{s_n : n > N\}, \end{align}

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 $$N+1 < i < k$$ 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

$\sigma_k \ge \left(1 - \frac NM\right)\inf\{s_n : n > N\}.$

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

$\inf\{\sigma_n : n > M\} \ge \left(1 - \frac NM\right)\inf\{s_n : n > N\}.$

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

$\liminf \sigma_n \ge \inf\{s_n : n > N\},$

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

$\liminf \sigma_n \ge \liminf s_n,$

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. $\sum{\frac {n^4}{2^n}}$
2. $\sum{ \frac{n!}{n^n} }$
3. $\sum{\frac {7}{3^n + 2n}}$
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.