If $b_n=a_n+a_{n+1}$ is converge to $L$. How can I prove that $\lim_{n\rightarrow\infty}\frac{a_n}{n}=0$ If $b_n=a_n+a_{n+1}$ is converge to $L\in\mathbb{R}$. How can I prove that $$\lim_{n\rightarrow\infty}\frac{a_n}{n}=0$$
 A: Let $\epsilon > 0$. We need to show that from some point onwards $|a_n| < n \epsilon$. We know that for large enough $n \geq k$ we have $$|a_n + a_{n+1} - L| < \frac{\epsilon}{2}.$$ It follows that $$|a_{n+2} - a_{n}| \leq |a_{n+2} + a_{n+1} - L| + |-a_{n+1} - a_n + L| < \epsilon.$$ 
For $n > k$, we have $|a_n| < \textrm{max}(|a_k|,|a_{k+1}|) + \frac{n-(k+1)}{2} \epsilon$. For $n$ very large this will be bounded by $n \epsilon$, as desired.
By the way, the sequence $a_n$ is not bounded in general. For example let $a_0 = 1, a_{2n} = a_{2n-2} + \frac{1}{n}$ and $a_{2n+1} = -a_{2n}$.
A: Note that $b_{n+1}-b_n = a_{n+2}-a_n$. It will be sufficient to show that
$$\lim_{n\to\infty}\frac{a_{2n}}{2n}=0$$
$$\lim_{n\to\infty}\frac{a_{2n+1}}{2n+1} = 0$$
Since
$$\lim_{n\to\infty}\frac{a_{2n+2}-a_{2n}}{2(n+1)-2n}=\lim_{n\to\infty}\frac{b_{2n+1}-b_{2n}}{2}=0$$
(both $b_{2n+1}$ and $b_{2n}$ tend to $L$), then by Stolz–Cesàro theorem the first one is proved. Analoguously the second one can be proved, since
$$\lim_{n\to\infty}\frac{a_{2n+3}-a_{2n+1}}{2(n+3)-(2n+1)}=\lim_{n\to\infty}\frac{b_{2n+2}-b_{2n+1}}{2}=0$$
A: Based on Arthur's answer, Here is an answer to my question.
Since $b_n$ converges to $L$, for any $\epsilon>0$ there is a $N\in\mathbb{N}$ such that
$n\geq N$ implies $|b_n-L|=|a_n+a_{n+1}-L|<\epsilon/2$. Then, for $n\geq N$, we have
\begin{align*}
|a_{n+2}-a_n|&\leq|a_{n+2}+a_{n+1}-L+L-a_{n+1}-a_{n}|\\
&\leq
|a_{n+2}+a_{n+1}-L|+|a_{n+1}+a_n-L|\\
&=\epsilon\,.\end{align*}
For a large enough $n>>N$, we get
\begin{align*}
|a_{n+2}-a_N|&=|a_{n+2}-a_n+a_n-a_{n-2}+a_{n-2}+a_{n-1}+\cdots+a_{N+2}-a_N|\\
&\leq |a_{n+2}-a_n|+|a_n-a_{n-2}|+\cdots+|a_{N+4}-a_{N+2}|+|a_{N+2}-a_N|\\
&\leq  |a_{N+n+2-N}-a_{N+n-N}|+|a_{N+n-N}-a_{N+n-2-N}|+\cdots+|a_{N+4}-a_{N+2}|+|a_{N+2}-a_N|\\
&< \epsilon\left(\frac{n+2-N}{2}\right)
\end{align*}
Rewriting the last inequality:
$$-\left(a_N+\epsilon\left(\frac{n+2-N}{2}\right)\right)\leq a_{n+2}\leq a_N+\epsilon\left(\frac{n+2-N}{2}\right)\,,$$
and
$$-\left(\frac{|a_N|}{n+2}+\frac{\epsilon N}{n+2}+\frac{\epsilon}{2}\right)\leq \frac{ a_{n+2}}{n+2}\leq \frac{|a_N|}{n+2}+\frac{\epsilon N}{n+2}+\frac{\epsilon}{2}\,.$$
By taking a limit of $n\rightarrow\infty$, since $a_N$ and  $N$ are fixed number, we get
$$-\frac{\epsilon}{2}<\lim_{n\rightarrow\infty}\frac{a_{n+2}}{n+2}<\frac{\epsilon}{2}\,.$$
Since the above inequality holds for any choice of $\epsilon,$ we get $$\lim_{n\rightarrow \infty}\frac{a_n}{n}=\lim_{n\rightarrow \infty}\frac{a_{n+2}}{n+2}=0\,.$$
