Let $\{x\_n\}\_{n=1}^{\infty}$ be a monotone decreasing sequence of positive real numbers such that $\sum\_{n=1}^{\infty} x\_n$ diverges. Also let $\{k\_n\}\_{n=1}^{\infty}$ be a strictly increasing sequence of positive integers such that $\sum\_{n=1}^{\infty} \frac{1}{k\_n}$ diverges. 
Can $\sum\_{n=1}^{\infty} x\_{k\_n}$ converge ?


Note :
If $\{k\_n\}\_{n=1}^{\infty}$ was linear (that is there were constants $A,B$ such that $k\_n = nA+B$ for all positive integers $n$) then the sum must diverge and this can be shown in an elementary way.
But, not all such sequences of positive integers are upper bounded by linear functions. For instance the sequence of consecutive prime numbers (since it is bounded below by $n ( \log n + \log \log n - 1)$).



Sum of a terms in a divergent series taken along indices the sum of whose reciprocal diverges. Can the sum converge?
Yes, this can happen. E.g., let $m\_1,m\_2,\dots$ be natural numbers such that 
$$m\_r\sim\ln r$$
(all asymptotic relations here are for $r\to\infty$). Let $k\_r:=m\_1+\dots+m\_r$, so that $k\_r\sim r\ln r$ and hence $\sum\_r1/k\_r=\infty$. Let 
\begin{equation}
 x\_n:=y\_r\sim1/(r\ln^2 r)\quad\text{if}\quad k\_{r-1}+1\le n\le k\_r 
\end{equation}
(with $k\_0=0$). Then 
$$\sum\_n x\_n=\sum\_r m\_ry\_r=\sum\_r\frac{1+o(1)}{r\ln r}=\infty.
$$
However,
\begin{equation}
 \sum\_r x\_{k\_r}=\sum\_r y\_r<\infty. 
\end{equation}


