# Thread: [SOLVED] Another lim inf/lim sup proof

1. ## [SOLVED] Another lim inf/lim sup proof

Let $(s_{n})$ be a sequence of nonnegative numbers, and for each n define $a_{n} = \frac{1}{n}(s_{1} +s_{2}+...+ s_{n})$. Show that $\lim \,inf \,s_{n} \le \lim \,inf \,a_{n} \le \lim \,sup \,a_{n} \le \lim \,sup \,s_{n}$. Also show that if $\lim s_{n}$ exists, then $\lim a_{n}$ exists and $\lim a_{n} = \lim s_{n}$.

For this one I am completely stuck. Any help would be appreciated.

2. Originally Posted by Pinkk
Let $(s_{n})$ be a sequence of nonnegative numbers, and for each n define $a_{n} = \frac{1}{n}(s_{1} +s_{2}+...+ s_{n})$. Show that $\lim \,inf \,s_{n} \le \lim \,inf \,a_{n} \le \lim \,sup \,a_{n} \le \lim \,sup \,s_{n}$. Also show that if $\lim s_{n}$ exists, then $\lim a_{n}$ exists and $\lim a_{n} = \lim s_{n}$.

For this one I am completely stuck. Any help would be appreciated.
What is the definition of $\limsup$?

3. $lim sup = \lim_{N\to \infty}\, sup\{s_{n} : n > N\}$

4. Any suggestions?

5. Apparently, for part of the proof I need to show that $M > N$ implies $sup\{a_{n} : n > M\} \le \frac{1}{M}(s_{1}+s_{2}+...+s_{N}) + sup\{s_{n} : n > N\}$ (and I don't even how to show that or why that's true...

6. Sorry for all the consecutive posts, but I am still absolutely stuck on this.