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.