Proving $\sum \frac{a_n}{n+x}$ converges uniformly on $[0,+\infty)$ Let $c_1, c_2,\cdots, c_m\in\Bbb R$ such that $c_1+c_2+\cdots+c_m=0$ and define $\{a_n\}$ so that $a_n=c_k$ if $n\equiv k \ \ (\mod m).$ 
How can we prove that $\displaystyle \sum_{k=1}^\infty\frac{a_n}{n+x}$ converges uniformly on $[0,\infty)$? I know I should use summation by parts, but I can't see how to proceed
 A: Since you were given the hint to use partial summation, why don't you? You have a sum of products $\displaystyle a_n\cdot\frac1{n+x}$, you transform it to a new sum of products of the sums of one sequence, and differences of the other one. Now summing the second factor doesn't make sense, so let $\displaystyle s_n=\sum^n_{k=1}a_k$. Assuming $s_0=0$, we have $a_k=s_k-s_{k-1}$ for $k\ge1$, so
\begin{align}\sum^n_{k=1}\frac{s_k-s_{k-1}}{k+x}&=\sum^n_{k=1}\frac{s_k}{k+x}-\sum^n_{k=1}\frac{s_{k-1}}{k+x}\\&=\sum^n_{k=1}\frac{s_k}{k+x}-\sum^{n-1}_{k=0}\frac{s_k}{k+1+x}\\&=\sum^n_{k=1}\frac{s_k}{(k+x)(k+1+x)}+\frac{s_n}{n+1+x}\end{align}
Now we know $a_{n+m}=a_n$, meaning $s_{n+m}=s_n+c_1+\ldots+c_m=s_n$, so the sequence $s_n$ is periodic, and thus bounded, $|s_n|\le S$ for all $n\ge1$. Now for $x\ge0$, we have $$\left|\frac{s_k}{(k+x)(k+1+x)}\right|\le\frac{S}{k(k+1)},$$ so the series $$\sum^\infty_{k=1}\frac{s_k}{(k+x)(k+1+x)}$$ converges uniformly due to the  majorant test of Weierstrass, and
$$\left|\frac{s_n}{n+x}\right|\le\frac{S}n\to0$$ as $n\to\infty$ uniformly in $x\ge0$, and that's what was to prove.
