Let {$a_n$} be a positive sequence of real numbers and $ \lim_{n \rightarrow \infty}a_n \rightarrow 0$. For any monotonic subsequence {$b_n$} of {$a_n$}

$\sum_{n = 1}^{\infty} b_n$ converges.

Does $\sum_{n = 1}^{\infty} a_n$ coverge?


For $n \geqslant 1$, let $m = \lfloor\sqrt{n}\rfloor$ and $k = n - m^2$. Let

$$a_n = a_{m^2+k} = \frac{1}{m^2} + \frac{k}{2m+1}\left(\frac{1}{m^2} - \frac{1}{(m+1)^2}\right).$$

We then have

$$a_n \geqslant \frac{1}{\lfloor n\rfloor^2},$$


$$\sum_{n=1}^{N^2-1} a_n = \sum_{m=1}^{N-1} \left(\sum_{k=0}^{2m} a_{m^2+k}\right) \geqslant \sum_{m=1}^{N-1} \frac{2m+1}{m^2} > 2\log N,$$

hence $\sum_{n=1}^\infty a_n = \infty$.

But since for $m \geqslant 1$ and $0 \leqslant k < 2m$ we have $a_{m^2+k} < a_{m^2+k+1}$, and $a_{m^2} > a_{(m+1)^2 + 2(m+1)} > a_{(m+1)^2}$, each monotonic subsequence can contain only one term between $a_{m^2}$ and $a_{(m+1)^2}$, hence the corresponding series converges (it's dominated by $\frac{2}{m^2}$).

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    $\begingroup$ Hi Dan, I'm sure you are right, but didn't follow how you got the inequality $\sum $ > 2/m. I can see that with an inequality we can collapse the $1/{m^2} - 1/{(m+1)^2}$, but that leaves me with a lot of $m^2$ terms in the denominator. Would you mind expanding it a bit for me? Thanks. $\endgroup$ – Betty Mock Sep 21 '13 at 17:04
  • $\begingroup$ I've explained a bit more. If it's not enough, ping me. $\endgroup$ – Daniel Fischer Sep 21 '13 at 19:51
  • $\begingroup$ Hi Daniel Fischer, I want to check the answer accepted, but can't until I understand it. Before I ask you more I want to take a little time to go over this and figure it out myself; and I'm up against a work deadline, so I can't get back to it right away. After 10/6 I'll have some time. In the meantime, could you just tell me if the funny characters around $\sqrt n$ mean "greatest integer", or what? $\endgroup$ – Betty Mock Sep 24 '13 at 0:11
  • $\begingroup$ Yes, those mean floor or greatest integer (Gauß brackets, if you're my age ;) The idea is to take a converging series and insert long slowly increasing parts between the terms, so that the entire series diverges, but a monotonic subsequence cannot contain more than one term per increasing stretch, so that a monotonic subsequence leads roughly back to the original convergent series. $\endgroup$ – Daniel Fischer Sep 24 '13 at 11:21
  • $\begingroup$ Hi Dan, yes, I've got it and thank you for your help. I studied math during the Pleistocene era; it's amazing I remember anything. The floor brackets are fine; we used something with a double vertical line so I wasn't quite sure. $\endgroup$ – Betty Mock Oct 7 '13 at 19:30

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