Convergence of the series $\sum_{n=1}^{\infty}\frac{x_n}{n^b}$

A slight modification of Dirichlet's test for convergence works for this problem:

Let $s_0=0$ and $s_n=x_1+\dots+x_n$ for $n\geq 1$, and choose a constant $C$ such that $|s_n|\leq Cn^a$ for all $n$. Then $$ \sum_{n=1}^N\frac{x_n}{n^b}=\sum_{n=1}^N\frac{s_n-s_{n-1}}{n^b}=\frac{S_N}{N^b}+\sum_{n=1}^{N-1}s_n[n^{-b}-(n+1)^{-b}] $$ by summation by parts.

For the first term, we have $$ \Big|\frac{S_N}{N^b}\Big|\leq \frac{C}{N^{b-a}}\to 0$$ as $N\to\infty$.

For the sum, since $n^{-b}>(n+1)^{-b}$ it follows that $$ \sum_{n=1}^{N-1}|s_n(n^{-b}-(n+1)^{-b})|=\sum_{n=1}^{N-1}|s_n|(n^{-b}-(n+1)^{-b})\leq C\sum_{n=1}^Nn^a(n^{-b}-(n+1)^{-b})=$$ $$=Cb\sum_{n=1}^{N-1}n^a\int_{n}^{n+1}x^{-b-1}\;dx\leq Cb\sum_{n=1}^{N-1}\int_n^{n+1}x^{-(b-a)-1}\;dx=Cb\int_1^{N}x^{-(b-a)-1}\;dx$$

Taking $N\to\infty$ and using the fact that $(b-a)+1>1$, we see that $$ \sum_{n=1}^{\infty}|s_n(n^{-b}-(n+1)^{-b})| $$ converges, and hence that $$ \sum_{n=1}^{\infty}s_n(n^{-b}-(n+1)^{-b}) $$ converges. Therefore $$ \sum_{n=1}^{\infty}\frac{x_n}{n^b}=\sum_{n=1}^{\infty}s_n(n^{-b}-(n+1)^{-b})$$ converges as well.