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