Prove that a summation is convergent

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.

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

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