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Let $\{x_n\}$ be a decreasing sequence of positive numbers and $\lim_{n \to \infty} x_n = 0$. Prove that the following summation is convergent.(Hint: Use alternating series test.)$$S = x_1 -\frac{1}{2}(x_1+x_2) +\frac{1}{4}(x_1+x_2+x_3)-\frac{1}{8}(x_1+x_2+x_3+x_4) + \dots$$

My try: Let $$S_n = \frac{1}{2^{n-1}}\sum_{m=1}^nx_m$$So we are interested in the convergence of $$S = \sum_{n=1}^{+\infty}(-1)^{n+1}S_n$$We can prove that $\{S_n\}$ is a decreasing sequence: $$S_{n+1} - S_n = \frac{1}{2^{n}}\sum_{m=1}^{n+1}x_m - \frac{1}{2^{n-1}}\sum_{m=1}^nx_m = \frac{x_{n+1}}{2^n} + (\sum_{m=1}^nx_m)(\frac{1}{2^n} - \frac{1}{2^{n-1}}) = \frac{1}{2^{n-1}}(\frac{x_{n+1}}{2} + (\sum_{m=1}^nx_m)(\frac{1}{2} - 1))$$ $$\frac{x_{n+1}}{2} + (\sum_{m=1}^nx_m)(\frac{1}{2} - 1)\le0 \iff x_{n+1} - \sum_{m=1}^nx_m \le 0 \iff x_{n+1} \le \sum_{m=1}^nx_m$$ Which is true since $x_{n+1} \le x_n$ and $x_n\ge 0$. Therefore we have $S_{n+1}\le S_n$. Obviously $S_n \ge 0$ for all $n$ and we should prove $\lim_{n\to \infty}S_n = 0$ in order to use alternating series test. I couldn't prove that and got stuck here.

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The series is absolutely convergent. Since $(x_n)$ is bounded $|x_1+x_2+..+x_n | \leq nM$ for some $M$ and $\sum \frac n {2^{n}}$ is convergent. Use Comparison Test to finish.

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  • $\begingroup$ $M=\sup_n |x_n|$ for example. $\endgroup$ – Tito Eliatron Dec 3 '20 at 8:56
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    $\begingroup$ I think we can simply set $M = x_1$. Is this correct? $\endgroup$ – S.H.W Dec 3 '20 at 9:01
  • $\begingroup$ @S.H.W Of course under the given conditions we can take $M=x_1$. But there is a lot of unnecessary hypotheses. So drop all the assumptions about $x_n$'s being positive decreasing and tending to $0$. Any bounded sequence $(x_n)$ will do. $\endgroup$ – Kavi Rama Murthy Dec 3 '20 at 9:06
  • $\begingroup$ @KaviRamaMurthy But we need some assumptions to prove $\{S_n\}$ is a decreasing sequence. $\endgroup$ – S.H.W Dec 3 '20 at 9:09
  • $\begingroup$ Do not use Alternating Series test. Just use comparison test. If $|a_n| \leq \frac n {2^{n}}$ the $\sum a_n$ is absolutely convergent. @S.H.W $\endgroup$ – Kavi Rama Murthy Dec 3 '20 at 9:11
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$$\lim_{n\to\infty}S_n=\lim_{n\to\infty}\frac{x_1+\ldots+x_n}{2^{n-1}} =\lim_{n\to\infty}\frac{x_1+\ldots+x_n}{n}\frac{n}{2^{n-1}}.$$

Now, The first factor is the arithmetic mean of $x_n$'s, so it converges also to $0$ (see here, for instance, but it is a classical application of Stolz Cesaro).

Morteover, it is straightforward that $\lim_{n\to\infty}\frac{n}{2^{n-1}}=0$. (If you do not want to calculate this limit, you can also argue that $0\le \frac{n}{2^{n-1}}\le 1$, so bounded).

So, $\lim_{n\to\infty}S_n=0$.

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  • $\begingroup$ Thanks. So it seems $\lim_{n \to \infty} x_n = 0$ is redundant and it's enough that $\lim_{n \to \infty} x_n = L$. $\endgroup$ – S.H.W Dec 3 '20 at 8:55
  • $\begingroup$ Yes, correct. I also think so. $\endgroup$ – Tito Eliatron Dec 3 '20 at 8:55

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