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? 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
$$|a_{n+2}-a_{n}-A|\le\varepsilon$$
 A: 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 $\lim\limits_{n\rightarrow\infty}{\frac{a_n-a_{n-1}}{n}}=0$
A: If $(c_N)_{N\geqslant 1}$ is a sequence of real numbers converging to $l$, then $N^{-1}\sum_{j=1}^Nc_j\to l$. With $c_N=a_{2N}$, we have $c_{n+1}-c_n\to 0$, hence $\frac{a_{2n}}{2n}\to A/2$. With $c_N=a_{2N+1}$, we obtain that $\frac{a_{2n+1}}{2n+1}\to A/2$. 
A: Hint: use the identities of the kind 
$$
d_n=
a_{n+1} -a_{n}
=
a_{n+1}-a_{n-1}+a_{n-2} -a_{n} +a_{n-1}-a_{n-2}
=
a_{n+1}-a_{n-1}+a_{n-2} -a_{n} + d_{n-2}
$$
Answers:  $\lim\limits_{n\rightarrow +\infty} a_n/n = A/2$ and  $\lim\limits_{n\rightarrow +\infty} (a_{n+1}-a_n)/n=0$
Example as a check: take $a_n=nA/2$.
