Proof of Sequence Let $a_n$ be a sequence on $\Bbb R$ and $a\in \Bbb R$ with $a_{n+1}-a_n\rightarrow a$. Show that $\frac{a_n}{n} \rightarrow a$
MY Attempt:
First, I layed out how one would usually go about it. Obviously, if $a_{n+1}-a_n\rightarrow a$, then $|a_{n+1}-a_n-a|<\epsilon$ for all $n\ge N$. Then I thought maybe i take the absolute value sign away maybe i'll see something.
$$-\epsilon+a<a_{n+1}-a_n<\epsilon+a$$ or
$$\frac{-\epsilon+a}{n}<\frac{a_{n+1}-a_n}{n}<\frac{\epsilon+a}{n}$$
Couldn't think of anything that would help...
Then i thought maybe another way:
$$\frac{a_n}{n}=\frac{a_N}{n}+\sum_{i=N}^{n-1}\frac{a_{N+1}-a_N}{n}$$
Then
$$\frac{a_n}{n}-a=\frac{a_N}{n}+\sum_{i=N}^{n-1}\frac{a_{N+1}-a_N}{n}-a$$
Then im stuck again....I didnt post this because its kind of embarrassing...
 A: Take an $\epsilon\gt0$ arbitrary, and since $a_{n+1}-a_n\to a$ for all $n\ge N$ for some $N\in\Bbb N$ you have, as you mentioned:
$$\frac{a-\epsilon}{n}\lt\frac{a_{n+1}-a_n}{n}\lt\frac{a+\epsilon}{n}$$
As you also mentioned, you can use telescoping series to find, for all $n\ge N$:
$$\frac{a_n}{n}=\frac{a_N}{n}+\sum_{k=N}^{n-1}\frac{a_{k+1}-a_k}{n}$$
All you had to do to complete the proof (Stolz-Cesaro does it immediately, but this one you can do more elementarily) is to use the above inequality and equality together:
$$\begin{align}\forall n(\in\Bbb N)\gt N:\quad\frac{a_N}{n}+\sum_{k=N}^{n-1}\frac{a-\epsilon}{n}&\lt\frac{a_n}{n}\lt\frac{a_N}{n}+\sum_{k=N}^{n-1}\frac{a+\epsilon}{n}\\\frac{a_N}{n}+\frac{1}{n}(n-N)(a-\epsilon)&\lt\frac{a_n}{n}\lt\frac{a_N}{n}+\frac{1}{n}(n-N)(a+\epsilon)\\\frac{1}{n}(a_N-N(a-\epsilon))+a-\epsilon&\lt\frac{a_n}{n}\lt a+\epsilon+\frac{1}{n}(a_N-N(a+\epsilon))\end{align}$$
The quantity $(a_N-N(a\pm\epsilon))$ is finite and fixed with respect to $n$; then if we let $n\to\infty$ on both sides we find that $\frac{1}{n}(a_N-N(a\pm\epsilon))\to0$:
$$a-\epsilon\le\liminf_{n\to\infty}\frac{a_n}{n}\le\limsup_{n\to\infty}\frac{a_n}{n}\le a+\epsilon$$
Since $\epsilon$ is here arbitrary, I may take limits as $\epsilon\to0^+$, and the squeeze theorem yields that: $$\lim_{n\to\infty}\frac{a_n}{n}=a$$
You had exactly the right idea!
