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.

Solution 1:

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 $s_n\to 0$