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}


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

  • $\begingroup$ 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. $\endgroup$ – adityaguharoy Jul 16 '18 at 14:59
  • $\begingroup$ @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.) $\endgroup$ – Steven Stadnicki Jul 16 '18 at 17:44
  • $\begingroup$ OK yes it was my bad. I appreciate the answer now. The accepted answer is quite related to this one. $\endgroup$ – adityaguharoy Jul 17 '18 at 12:17

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