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

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

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}
• I don't understand how you get the "subseries behaves like" $n ( \log n \cdot \log n)$. The sum does not behave like that. And even if you meant that the terms behave like that still it is wrong. Jul 16, 2018 at 14:59
• @adityaguharoy It's written a little sloppily but this basic argument is correct: if $f(x) = x\log x$, $f(x\log x)=(x\log x)\cdot\log(x\log x)) = (x\log x)(\log x+\log\log x) \gt x\log^2x$; so you can take $x_n=\frac1{x\log x}$ and $k_n\approx x\log x$ in your question. (Either floor or ceiling would work fine there; the differences aren't enough to affect convergence.) Jul 16, 2018 at 17:44