Sum of a terms in a divergent series taken along indices the sum of whose reciprocal diverges. Can the sum converge?

Question

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 $n ( \log n + \log \log n - 1)$).

Answer 1

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}

Answer 2

What about if (for n bigger than 2) you take x(n)=(nlog(n))^(-1) and k(n) roughly same as x(n)^(-1) (say integral part of nlog(n))?

Then both conditions required are satisfied, while the subseries wanted behaves like (n*log(n)*log(n))^(-1) which is convergent