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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.

  • Looks awfully a lot like this problem, but I’m guessing that the two problems are unrelated. – Berrick Caleb Fillmore Apr 25 '15 at 08:17
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    $s_n=(-2)^n$,then $a_n=(-2)^n+2(-2)^{n-1}=0$,but $s_n$ is not convergent.. – Golbez Apr 25 '15 at 08:23
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    @BerrickFillmore Indeed not the same problem, since one is correct and the conclusion of the other is wrong. – Did Apr 25 '15 at 08:36
  • @Golbez My source might be wrong http://prac.im.pwr.wroc.pl/~kwasnicki/pl/stuff/tbb-hyper.pdf (page 76, Q 2.14.17). But what if the constant sequences are not allowed. –  Apr 25 '15 at 08:41
  • @Did: Thanks for the notice! – Berrick Caleb Fillmore Apr 25 '15 at 08:49
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    @Umakant You know what? The conditions $a_n = s_n + 2s_{n\color{red}{-1}}$ and $a_n = s_n + 2s_{n\color{red}{+1}}$ are not equivalent. – Did Apr 25 '15 at 08:54
  • @Umakant Actually, we may have$$s_n=\sum_{k=0}^{n-1} (-2)^k a_{n-k}+(-2)^n s_0.$$ For instance, we let $a_n=2^{-n}$. Then $$s_n=2^{-n}\frac{(-4)^n-1}{-5}+(-2)^{s_0}.$$ When $s_0\not=\frac{1}{5}$, the sequence will diverge. – Golbez Apr 25 '15 at 08:56
  • @Golbez The mystery is solved, please see my previous comment. – Did Apr 25 '15 at 08:56
  • @Did Sorry for the typo. Just edited it. –  Apr 25 '15 at 08:57
  • And now, duplicates, near-duplicates, functional duplicates of the question abound. – Did Apr 25 '15 at 09:05
  • @Umakant: Notice that $$ \forall n \in \mathbb{N}{0}: \quad s{n + 1} = \sum_{k = 0}^{n} \frac{1}{2} \left( - \frac{1}{2} \right)^{k} a_{n - k} + \left( - \frac{1}{2} \right)^{n + 1} s_{0}. $$ – Berrick Caleb Fillmore Apr 25 '15 at 09:16
  • @BerrickFillmore Thanks for the relation, but can I know the intuition behind reaching it. –  Apr 25 '15 at 09:26
  • Write $ s_{n + 1} $ in terms of $ a_{n} $ and $ s_{n} $. Then $ s_{n} $ itself can be written in terms of $ a_{n - 1} $ and $ s_{n - 1} $. Keep doing this until you hit $ s_{0} $. – Berrick Caleb Fillmore Apr 25 '15 at 09:28
  • @BerrickFillmore Thanks, that is all I wanted. –  Apr 25 '15 at 09:30

3 Answers3

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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 $s_n\to 0$

Golbez
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  • Yes that’s correct. I also was thinking in the lines of relating $s_n$ to $a_n$'s which would make the proof simpler. All I want to know is the intuition behind guessing that relation (in this case your first equation). –  Apr 25 '15 at 09:25
  • Okay, it's simple when you just compute the first three terms. $a_0=s_0+2s_1$ thus $s_1=\frac{1}{2}(a_0-s_0)$ and $s_2=\frac{1}{2}(a_1-s_1)=\frac{1}{2}a_1-\frac{1}{4}a_0+\frac{1}{4}s_0$... Finding the pattern will give you the answer, at least in this simple case – Golbez Apr 25 '15 at 09:31
  • Thanks, that was all I wanted. Should have figured it out. –  Apr 25 '15 at 09:32
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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: $$s_{n+1}=\frac{1}{3}\lim_{n\rightarrow \infty}y_i$$

Thomas A. Anderson
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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.

Gabriel Romon
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