Prove $\{s_n\}$ converges if $\{a_n = s_n + 2s_{n+1}\}$ converges. Prove that $\{s_n\}$ is convergent if $\{a_n\}$ is convergent where $a_n = s_n + 2s_{n+1}$.
This is an old (1950) Putnam question. 
Clearly $s_n + 2s_{n+1} \rightarrow L$. It looks obvious that $s_n \rightarrow L/3$, but how to prove it.
Here is my exact problem. For sufficiently large $n$ all of $s_n + 2s_{n+1}$, $s_{n+1}+2s_{n+2}$, $s_{n+2}+2s_{n+3}$ $\ldots$ are all nearly equal. How do I derive that $s_n$, $s_{n+1}$, $s_{n+2} \ldots$ are also nearly same (with a mathematical argument). There must be a simple trick here which eludes me.
Any hints are welcome.
 A: We may prove by induction that
$$s_n=-\sum_{k=1}^{n} \left(-\frac{1}{2}\right)^k a_{n-k}+\frac{s_0}{(-2)^n} $$
WLOG we may assume $a_{n}\to 0$. It suffices to prove $$\sum_{k=1}^{n} \left(-\frac{1}{2}\right)^k a_{n-k}\to 0$$Now let $M=\sup|a_n|$. And for all $\epsilon>0$, exists $N$ such that $|a_n|<\epsilon$ when $n-\sqrt{n}>N$.The summation $$\sum_{k=1}^{[\sqrt{n}]} \left(-\frac{1}{2}\right)^k a_{n-k}+\sum_{k=[\sqrt{n}]+1}^{n} \left(-\frac{1}{2}\right)^k a_{n-k}.$$
The first part$$\left|\sum_{k=1}^{[\sqrt{n}]} \left(-\frac{1}{2}\right)^k a_{n-k}\right|\le \epsilon\sum_{k=1}^{[\sqrt{n}]} \left(\frac{1}{2}\right)^k\le C_1 \epsilon$$
The second part$$\left|\sum_{k=[\sqrt{n}]+1}^{n} \left(-\frac{1}{2}\right)^k a_{n-k}\right|\le M\frac{C}{2^{\sqrt{n}}}.$$
Thus in all $s_n\to 0$
A: Notice that
$$s_{n+1}=\frac{1}{2}\sum_{i=0}^n(-\frac{1}{2})^{n-i}a_i$$
Let $t_{n,i}=(-\frac{1}{2})^{n-i}$，Notice that:
$$\lim_{n\rightarrow \infty}\sum_{i=0}^nt_{n,i} = \frac{2}{3}$$
$$\forall k,\lim_{n\rightarrow \infty}t_{n,k}= 0$$
Use Toeplitz's theorem. 
Then:
$$\lim_{n\rightarrow \infty}\sum_{i=0}^nt_{n,i}y_i = \frac{2}{3} \lim_{n\rightarrow \infty}y_i$$
Then:
$$s_{n+1}=\frac{1}{3}\lim_{n\rightarrow \infty}y_i$$
A: Let $L = \lim_n a_n$. We prove that $(s_n)$ converges to $L/3$. It is natural to look for some upper bound of $|s_{n+1}-L/3|$ involving $|s_n-L/3|$ and $|a_n-L|$.
Indeed $|s_{n+1}-L/3| = |(a_n-s_n)/2 - L/3| \leq 1/2 |a_n-L| + 1/2 |s_n-L/3|$. Iterating, we find
$$|s_{n+1}-L/3|\leq \frac 1{2^{n+1}}|s_0- L/3| +  \sum_{k=0}^n \frac 1{2^{k+1}}|a_{n-k}-L|,$$
so it suffices to prove that
$$\sum_{k=0}^n \frac 1{2^{k+1}}|a_{n-k}-L|\xrightarrow[]{n\to \infty } 0.$$
Let $\epsilon>0$. There is some $N\geq 0$ such that $n\geq N\implies |a_n-L|\leq \epsilon/2$.
For $n\geq N$ we have the bound
$$
\begin{align}
\sum_{k=0}^n \frac 1{2^{k+1}}|a_{n-k}-L|
&\leq \frac \epsilon 2 \sum_{k=0}^{n-N} \frac 1{2^{k+1}} + \max(|a_0-L|,\ldots,|a_{N-1}-L|)\sum_{k=n-N+1}^n \frac 1{2^{k+1}} \\
&\leq \frac \epsilon 2 + \max(|a_0-L|,\ldots,|a_{N-1}-L|) \frac 1{2^{n-N+1}}
\end{align}$$
The second summand in the RHS goes to $0$ and the proof is easily finished.
