Absolute and Uniform convergence of series of function Question
Given
$$\sum_{n=1}^\infty \frac {a_n}{n^x}$$
when $x=s$ the series converges.
Prove that:


*

*It converges uniformly in $[s,\infty)$

*It converges absolutely in $(s+1,\infty)$
Thoughts


*

*Bounding it from above with $\sum_{n=1}^\infty \frac {a_n}{n^s}$ and using M-test.

*Bounding it from above with $\sum_{n=1}^\infty \frac {a_n}{n^{s+1}}$ and splitting that to $\frac 1n$ anf $\frac {a_n}{n^s}$, and proving convergence of that using Dirichlet.
Are these ways good? Any hints? We feel not sure about it.
 A: The second part is not hard. Since the terms $a_n/n^s$ are bounded by some $M$ (otherwise they would not form a convergent series), we have 
$$\left|\frac{a_n}{n^x}\right| \le \frac{M}{b^{x-s}}$$
and the series on the right converges when $x-s>1$.
The first part is more subtle.There is a detailed proof in Uniform convergence of Dirichlet series, though for a somewhat different setup. Here is how it works. Let $$R_n=\sum_{k>n}\frac{a_n}{n^s}$$ which is the remainder of a convergent series, hence $R_n\to 0$ as $n\to\infty$. Write the series as $\sum (R_{n-1}-R_n)/n^{x-s}$ and use summation by parts. Up to irrelevant boundary terms, summation by parts turns the series to 
$$\sum \left(\frac{1}{(n+1)^{x-s}}-\frac{1}{n^{x-s}}\right) R^n \tag{1}$$
The tail ($n\ge N$) of series (1) can be estimate uniformly in $x$: it does not exceed
$$\max_{n\ge N } |R_n|  \sum_{n\ge N} \left(\frac{1}{n^{x-s}}-\frac{1}{(n+1)^{x-s}}\right) \le 
\max_{n\ge N } |R_n| \to 0
$$
because the last series telescopes to $1/N^{x-s}\le 1$.
