Convergence of $\frac{1}{1^b}+\frac{1}{2^b}+\frac{a}{3^b}+\cdots+\frac{1}{(3k-2)^b}\frac{1}{(3k-1)^b}\frac{a}{(3k)^b}+\cdots$ For which values of $a,b\in\mathbb{R}$ the series:
$$\frac{1}{1^b}+\frac{1}{2^b}+\frac{a}{3^b}+\cdots+\frac{1}{(3k-2)^b}\frac{1}{(3k-1)^b}\frac{a}{(3k)^b}+\cdots$$
converges?
Attempt:
$$\frac{1}{1^b}+\frac{1}{2^b}+\frac{a}{3^b}+\cdots+\frac{1}{(3k-2)^b}\frac{1}{(3k-1)^b}\frac{a}{(3k)^b}+\cdots=\sum_{3k-2,3k-1}\frac{1}{n^b}+\sum_{3k}\frac{a}{n^b}$$
So, maybe, $b>1$ and $a\in\mathbb{R}$?
 A: We do indeed have convergence if $b\gt 1$. For let $c=\max(1,|a|)$. Then the $n$-th term of our series has absolute value $\lt \frac{c}{n^b}$. So by comparison, the series converges absolutely, and therefore converges. 
We have divergence for $b\le 0$, since the terms do not have limit $0$.
So we only need to worry about $0\lt b\le 1$. In that case, if  $a\ne -2$, we have divergence. We give a brief sketch of a proof.  Call our series $\sum_1^\infty x_k$. 
Note that $\lim_{k\to\infty}\frac{(3k+1)^b}{(3k+3)^b}=\lim_{k\to\infty}\frac{(3k+2)^b}{(3k+3)^b}=1$. So by taking $k$ large enough, we can make both these ratios as close to $1$ as we please. 
Suppose, for example, that $b+2\gt 0$. Let $b+2=c$. Then by taking $k$ large enough, we can make the  "sum" of the the $3$ consecutive terms $x_{3k+1}+x_{3k+2}+x_{3k}$ greater  than $\dfrac{c/2}{(3k)^b}$. Thus we have divergence by a $p$-series comparison. 
If $b+2\lt 0$, let $c=-(b+2)$. Essentially the same argument as the one for positive $b+2$ works. 
We have now dealt with all cases except $0\lt b\le 1$. In that case, the term $\frac{a}{(3k)^b}=-\frac{2}{(3k)^b}$ provides enough cancellation for convergence. We first deal with the simplest case, $b=1$.  In that case, the three-term sum $x_{3k+1}+x_{3k+2}+x_{3k}$ simplifies to
$$\frac{9k+4}{(3k+1)(3k+2)(3k+3)}.$$
The sum of three-term sums converges by Comparison. And since the $x_n$ have limit $0$, this implies that $\sum x_n$ converges.  
For $0\lt b\lt 1$, essentially the same idea works. We need to estimate 
$x_{3k+1}+x_{3k+2}+x_{3k}$.  Bring to a common denominator as in the case $b=1$.  Next  use the fact that for $i=1,2,3$ we have
$$(3k+i)^b =(3k)^b\left(1+\frac{i}{3k}\right)^b =(3k)^b\left(1+\frac{ib}{3k}+o(1/k)\right).$$
There is cancellation of "main" terms, and we end up concluding that for some positive constant $c$, if $k$ is large enough then $0\lt x_{3k+1}+x_{3k+2}+x_{3k}\lt \frac{1}{(3k)^b}\cdot \frac{c}{k}$. This is enough to prove convergence of $\sum_k (x_{3k+1}+x_{3k+2}+x_{3k})$, and therefore of the original series. 
Summary:  We have convergence for arbitrary $a$ and $b\gt 1$. We also have convergence if $a=-2$ and $0\lt b\le 1$.  In all other cases we have divergence.  
