Prove that a summation is convergent

Question

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.

Answer 1

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.

Answer 2

$$\lim_{n\to\infty}S_n=\lim_{n\to\infty}\frac{x_1+\ldots+x_n}{2^{n-1}} =\lim_{n\to\infty}\frac{x_1+\ldots+x_n}{n}\frac{n}{2^{n-1}}.$$

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