sequences and series - Show that if $sum_{k=1}^m c_k =0$, $sum_{n=0}^{infty} sum_{k=1}^m frac{c_k}{nm+k}$ converges.

This is a generalization of this:
Is this:$\sum_{n=1}^{\infty}{(-1)}^{\frac{n(n-1)}{2}}\frac{1}{n}$ a convergent series?

Here is my solution.

To show that
if
$\sum_{k=1}^m c_k =0$,
$\sum_{n=0}^{\infty} \sum_{k=1}^m \frac{c_k}{nm+k}$
converges,

let
$s(n) =\sum_{k=1}^m \frac{c_k}{nm+k}$.
Then

$\begin{array}\\ s(n)-\sum_{k=1}^m \frac{c_k}{nm} &=\sum_{k=1}^m \frac{c_k}{nm+k}-\sum_{k=1}^m \frac{c_k}{nm}\\ &=\sum_{k=1}^m c_k(\frac1{nm+k}-\frac1{nm})\\ &=\sum_{k=1}^m c_k\frac{nm-(nm+k)}{(nm+k)nm}\\ &=-\sum_{k=1}^m c_k\frac{k}{(nm+k)nm}\\ &=-\frac1{n^2}\sum_{k=1}^m c_k\frac{k}{(m+k/n)m}\\ \end{array}$

Since
$\sum_{k=1}^m \frac{c_k}{nm} =\frac1{nm}\sum_{k=1}^m c_k =0$,

$\begin{array}\\ |s(n)| &=|\frac1{n^2}\sum_{k=1}^m c_k\frac{k}{(m+k/n)m}|\\ &\le\frac1{n^2}\sum_{k=1}^m \big|c_k\frac{k}{(m+k/n)m}\big|\\ &<\frac1{n^2}\sum_{k=1}^m \big|c_k\frac{k}{(m)m}\big|\\ &=\frac1{n^2m^2}\sum_{k=1}^m \big|kc_k\big|\\ &=\frac{C}{n^2m^2}\\ \end{array}$

where
$C = \sum_{k=1}^m \big|kc_k\big|$.

Therefore
$\sum_{n=1}^{\infty} s(n)$

is absolutely convergent.

Answer

Another way can be this. Let $A_{k}=\sum_{s\leq k}c_{s}$. We have by partial summation

$$\sum_{k\leq m}\frac{c_{k}}{nm+k}=\sum_{k\leq m-1}\frac{A_{k}}{\left(nm+k\right)\left(nm+k+1\right)}=\frac{1}{n^{2}}\sum_{k\leq m-1}\frac{A_{k}}{\left(m+k/n\right)\left(m+\left(k+1\right)/n\right)}$$ then $$\left|\sum_{k\leq m}\frac{c_{k}}{nm+k}\right|\leq\frac{1}{n^{2}}\sum_{k\leq m-1}\frac{\left|A_{k}\right|}{\left(m+k/n\right)\left(m+\left(k+1\right)/n\right)}\leq\frac{1}{n^{2}m^{2}}\sum_{k\leq m-1}\left|A_{k}\right|=$$ $$=\frac{C}{n^{2}m^{2}}.$$