TITLE: How to prove these two limits $\lim_{n\to\infty}\frac{a_{n}}{n}$ and $\lim_{n\to\infty}\frac{a_{n+1}-a_{n}}{n}$ exist?
QUESTION [6 upvotes]: Assume that $$\lim_{n\to\infty}(a_{n+2}-a_{n})=A$$
show that
$\displaystyle \lim_{n\to\infty}\dfrac{a_{n}}{n}$and $\displaystyle\lim_{n\to\infty}\dfrac{a_{n+1}-a_{n}}{n}$ exist and find these limits.
maybe this problem have some methods, Thank you.
since
 $$\lim_{n\to\infty}(a_{n+2}-a_{n})=A$$
so there exists $N$, and for $\forall \varepsilon>0$, such
$$|a_{n+2}-a_{n}-A|\le\varepsilon$$

REPLY [2 votes]: It is given that for every $\varepsilon>0$ there exist a $N_0$ such that $n>N_0\implies$ $|a_{n+2}-a_n-A|<\varepsilon\implies a_n+A-\varepsilon<a_{n+2}<a_n+A+\varepsilon $
$a_{N'}+\frac{(n-2-N')}{2}A-\frac{(n-2-N')}{2}\varepsilon <a_{n}<a_{N'}+\frac{(n-2-N')}
{2}A+\frac{(n-2-N')}{2}\varepsilon \implies\\ \frac{a_{N'}}{n}+\frac{A}{2}-\frac{(2+N')}{2n}\cdot A-\varepsilon+\frac{(2+N')}{2n}\varepsilon<\frac{a_n}{n}<\frac{a_{N'}}{n}+\frac{A}{2}-\frac{(2+N')}{2n}\cdot A+\varepsilon-\frac{(2+N')}{2n}\varepsilon \implies \\ \boxed{\lim\limits_{n\rightarrow \infty}{\frac{a_n}{n}}=\frac{A}{2}}$
Where $N'$ is $N_0$ when both $n$ and $N_0$ have same parity and $N_0+1$ otherwise.
Similarly, forming upper and lower bounds for $a_n-a_{n-1}$  we can arrive at $\lim\limits_{n\rightarrow\infty}{\frac{a_n-a_{n-1}}{n}}=0$