sequences and series - If $sum_{k=0}^{r-1} c_k =0 $, and $a_n to 0$, does $sum_{n=0}^{infty} sum_{k=0}^{r-1} c_ka_{nr+k} $ converge?

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

Here is my question:

If

$\sum_{k=0}^{r-1} c_k
=0
$,
and
$a_n$ is a
monotonic decreasing series
such that
$a_n \to 0$,
does
$\sum_{n=0}^{\infty}c_{n \bmod r}a_n

=\sum_{n=0}^{\infty} \sum_{k=0}^{r-1} c_ka_{nr+k}
$
converge?

If $a_n = \frac1{n+1}$,
then the sum does converge
as shown by my question here:
Show that if $\sum_{k=1}^m c_k =0 $, $\sum_{n=0}^{\infty} \sum_{k=1}^m \frac{c_k}{nm+k} $ converges.

I can show that

the sum does converge
as long as the
$a_n$ satisfies
some smoothness properties:

Assume that
$a_n
=f(n)
$,
where

$f(x)$ is differentiable,
$f(x) \to 0$,
$f'(x)
\to 0
$,
and
$|f''(x)|
< C/x^2
$
for some $C > 0$.

Let
$s(n)
=\sum_{k=0}^{r-1} c_ka_{nr+k}
$,
and
$S(n)
=\sum_{j=0}^n s(j)
=\sum_{n=0}^{nr+r-1} a_n
$.

Then

$\begin{array}\\
s(n)-\sum_{k=0}^{r-1} c_ka_{nr}
&=\sum_{k=0}^{r-1} c_ka_{nr+k}-\sum_{k=0}^{r-1} c_ka_{nr}\\
&=\sum_{k=0}^{r-1} c_k(a_{nr+k}-a_{nr})\\
&=\sum_{k=0}^{r-1} c_k(f(nr+k)-f(nr))\\
&=\sum_{k=0}^{r-1} c_k(kf'(nr)+(k^2/2)f''(nr+tk))
\quad\text{where }0 \le t \le 1\\

&=\sum_{k=0}^{r-1} c_kkf'(nr)+\sum_{k=0}^{r-1}c_k(k^2/2)f''(nr+tk)\\
&=f'(nr)\sum_{k=0}^{r-1} c_kk+\sum_{k=0}^{r-1}c_k(k^2/2)f''(nr+tk)\\
&=f'(nr)S+s_1(n)\\
\end{array}
$

where
$S = \sum_{k=0}^{r-1} c_kk
$
and

$s_1(n)
=\sum_{k=0}^{r-1}c_k(k^2/2)f''(nr+tk)
$.

Since
$\sum_{k=0}^{r-1} c_ka_{nr}
=a_{nr}\sum_{k=0}^{r-1} c_k
=0
$,
$s(n)

=f'(nr)S+s_1(n)
$.

Also,

$\begin{array}\\
|s_1(n)|
&=|\sum_{k=0}^{r-1}c_k(k^2/2)f''(nr+tk)|\\
&\le\sum_{k=0}^{r-1}|c_k(k^2/2)f''(nr+tk)|\\
&\le\sum_{k=0}^{r-1}|c_k(k^2/2)\frac{C}{(nr+tk)^2}|\\

&\le\sum_{k=0}^{r-1}|c_k(k^2/2)\frac{C}{(nr)^2}|\\
&=\frac{C}{2(nr)^2}\sum_{k=0}^{r-1}|c_kk^2|\\
&=\frac{CC_1}{2(nr)^2}
\quad\text{ where } C_1 =\sum_{k=0}^{r-1}|c_kk^2|\\
\end{array}
$

Therefore,
$\sum |s_1(n)|
$

converges.

If $m < n$,

$\begin{array}\\
S(n)-S(m)
&=\sum_{j=m+1}^n s(j)\\
&=\sum_{j=m+1}^n (f'(jr)S+s_1(j))\\
&=S\sum_{j=m+1}^n f'(jr)+\sum_{j=m+1}^ns_1(j)\\
\end{array}

$

By the smoothness assumption
on $f(x)$,
$f'(jr)
\approx\frac1{r}\int_{jr}^{jr+r} f'(t)dt
$,
so that

$\begin{array}\\

\sum_{j=m+1}^n f'(jr)
&\approx\sum_{j=m+1}^n \frac1{r}\int_{jr}^{jr+r} f'(t)dt\\
&= \frac1{r}\int_{(m+1)r}^{nr+r} f'(t)dt\\
&= \frac1{r}(-f((m+1)r)+f(nr+r))\\
\end{array}
$

Since
$f(x) \to 0$
as $x \to \infty$,

$f(nr+r)-f((m+1)r)
\to 0
$
as $m$ and $n \to \infty$.

Since
$\sum |s_1(n)|$
converges,
$\sum_{j=m+1}^n s_1(j)
\to 0$

as $m$ and $n \to \infty$.

Therefore,
$S(n)-S(m)
\to 0$
as $m$ and $n \to \infty$,
so
$\lim_{n \to \infty} S(n)
$
exists.

Answer

The partial sums $C_N$ of the sequence $c_{n\text{ mod } r}$ satisfy $$|C_N| = \left|\sum_{n=0}^Nc_{n\text{ mod }{r}}\right| \leq \sum_{n=0}^{r-1}|c_{n}|$$ and is therefore bounded. Since $a_{n}$ is monotonicly decreasing with $\lim\limits_{n\to\infty} a_n = 0$ we have that Dirichlet's test apply and it follows that the series $\sum c_{n\text{ mod }r}a_n$ converges.