Does anybody know how to show analytically the convergence (divergence ?) of $\enspace\displaystyle \sum\limits_{n=1}^\infty \frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{\sqrt{n}}\enspace$ or some literature for it ? (Thanks!)

Note: Unfortunately I am still missing a useful approach.


I want to thank all here for the nice help.

Result: The series is divergent (but limited) as it is written.

It's possible to change to $\enspace\displaystyle\sum\limits_{n=1}^\infty (-1)^n (-2+\sum\limits_{k=n^2}^{(n+1)^2-1}\frac{1}{\sqrt{k}})\enspace$ to avoid the oscillation and to get convergence.

  • $\begingroup$ Does it converge? I think it doesn't. $\endgroup$ – EUNSUNG LIM Apr 7 '17 at 13:53
  • $\begingroup$ @EUNSUNGLIM LIM : I think so by numerically tests ... but maybe my possibilites are not good enough. $\endgroup$ – user90369 Apr 7 '17 at 13:56
  • $\begingroup$ $ \sum_{n=1}^{\infty} \frac{(-1)^{[\sqrt{n}]}}{\sqrt n} $= $-1-\frac{1}{\sqrt 2} -\frac{1}{\sqrt 3}+\frac{1}{\sqrt 4}+\frac{1}{\sqrt 5}+\frac{1}{\sqrt 6}+\frac{1}{\sqrt 7}+\frac{1}{\sqrt 8} +........ $ . This is an alternating series and hence it is convegent. Because each successive term is numerically less than the preceding term and $ \lim_{n \rightarrow \infty } \frac{1}{\sqrt n} \rightarrow 0 $ $\endgroup$ – M. A. SARKAR Apr 7 '17 at 14:18
  • $\begingroup$ @mabmath : Thanks, yes (by Leibniz if the terms go to 0) but each successive term is numerically less than the preceding term still has to be proofed (I mean: I still have to proof this). $\endgroup$ – user90369 Apr 7 '17 at 14:34
  • $\begingroup$ @mabmath, this is not an alternating series. The signs in an alternating series must go back and forth at each step. (That's what "alternating" means.). You are possibly thinking of Dirichlet's test. But even there, the running sum of the signs must stay bounded, which is not the case here. As several answers show, the OP's sum diverges. $\endgroup$ – Barry Cipra Apr 7 '17 at 15:02

You can prove that it diverges by observing that the sum of each "block" is greater than a constant, that is, $$\sum_{k=n^2}^{(n+1)^2-1}\frac{1}{\sqrt{k}}>C$$for some positive $C$.

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    $\begingroup$ Thanks! Unfortunately, this is an alternating set of estimates, upper and lower sub-values ​​are still too inaccurate here - perhaps there are appropriate estimates. I will try. $\endgroup$ – user90369 Apr 7 '17 at 14:05
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    $\begingroup$ The LHS is greater than $(2n+1$) times the smallest term in the sum. So the LHS is greater than $(2n+1)/\sqrt {n^2+2n}\;$ which is greater than $(2n+1)/(n+1)$ which $\to 2$ as $n\to \infty$.................+1. $\endgroup$ – DanielWainfleet Apr 7 '17 at 14:14
  • $\begingroup$ To the OP; Why do you say the estimate is too inaccurate? The fact that the same $C>0$ exists for all $n$ is enough too guarantee divergence. $\endgroup$ – DanielWainfleet Apr 7 '17 at 14:18
  • $\begingroup$ This is right. such C exists $\endgroup$ – EUNSUNG LIM Apr 7 '17 at 14:20
  • $\begingroup$ @user254665 : $s_n:=\sum\limits_{k=n^2}^{(n+1)^2-1}\frac{1}{\sqrt{k}}$ ; If it's $s_n>s_{n+1}$ then the convergence is clear because of Leibniz. But I still haven't proofed this. $\endgroup$ – user90369 Apr 7 '17 at 14:32

Sorry, I have forgotten how to type in LateX, so here it is by hand.

Basically, this is an explicit version of the other answer.

Essentially, we show that each block has a minimal magnitude (infimum) by which it add or subtracts from the partial sum. This non-decreasing (constant) bound makes the series fail the criteria for the alternating series test (I believe)

enter image description here

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Interestingly, if we study the blocks of partial terms, they tend towards "$2$" in magnitude, so the series diverges more "hectically" as we take $n$ to infinity than our bound may suggest.

  • $\begingroup$ Thanks! But because of technical reasons I cannot see images at the moment. $\endgroup$ – user90369 Apr 7 '17 at 14:41
  • $\begingroup$ Letting the "epsilons" go to $0$ represents an "even if" scenario. Indeed the sequence of epsilons on the second page, I believe, actually tends to $2-(2/3^{0.5})$ $\endgroup$ – Just_a_fool Apr 7 '17 at 14:43
  • $\begingroup$ I would use (absolute value of partial sum)$\ge (2n+1)/(n+1)>1$, it is much simpler. By the way, one can prove that every partial sum is greater than $2$ using simple integral. $\endgroup$ – didgogns Apr 7 '17 at 14:45
  • $\begingroup$ @didgogns : Thanks for the hint, than I have a problem with these partial sums. Do you know what the limit is for the partial sum ($s_n$) for $n\to\infty$ ? Then maybe it's possible to create a Leibniz series by substracting this limit from each $s_n$ . $\endgroup$ – user90369 Apr 7 '17 at 14:51
  • $\begingroup$ @user90369 Sorry about that. I would type it up if I knew how to. Also, didgogns Your idea is better (to simplify it a bit more to begin with). You mean you look at Ln((f(x)) as a bound and factor out (subtract) something large enough? $\endgroup$ – Just_a_fool Apr 7 '17 at 14:53

This does not converge; it oscillates by approximately $2$.

If $m^2 \le m \le m^2+2n $, then $\lfloor \sqrt{n}\rfloor = m$.


$\begin{array}\\ S(M) &=\sum_{n=1}^{M^2+2M} \frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{\sqrt{n}}\\ &=\sum_{m=1}^{M} \sum_{n=m^2}^{m^2+2m} \frac{(-1)^m}{\sqrt{n}}\\ &=\sum_{m=1}^{M} (-1)^m\sum_{n=m^2}^{m^2+2m} \frac{1}{\sqrt{n}}\\ &=\sum_{m=1}^{M} (-1)^ms(m) \qquad\text{where }s(m)=\sum_{n=m^2}^{m^2+2m} \frac{1}{\sqrt{n}}\\ \end{array} $

$s(m) \lt \frac{2m+1}{m} = 2+\frac1{m} $ and $s(m) \gt \frac{2m+1}{m+1} =2-\frac1{m+1} $.

Therefore, as $n$ goes from $m^2$ to $m^2+2n$, the value of the sum changes by approximately $2$, increasing when $n$ is even and decreasing when $n$ is odd.

Therefore the series does not converge, since it oscillates.

However, there might be a sense in which the series converges since we can write $S(2M) =\sum_{m=1}^{2M}(-1)^ms(m) =\sum_{m=1}^{M}(s(2m)-s(2m-1)) $.

I can show that that $s(m)-s(m+1) \to 0$, but I can not yet show that $\lim_{M \to \infty}\sum_{m=1}^{M}(s(2m)-s(2m-1)) $ exists.

Here's what I have so far, and I will leave it at this:

$\begin{array}\\ s(m)-s(m+1) &\gt (2-\frac1{m+1})-(2+\frac1{m+1})\\ &= -\frac{2}{m+1}\\ \text{and}\\ s(m)-s(m+1) &\lt (2+\frac1{m})-(2-\frac1{m+2})\\ &= \frac1{m}+\frac1{m+2}\\ &= \frac{2m+2}{m(m+2)}\\ \text{so}\\ s(m)-s(m-1) &= O(\frac1{m})\\ \end{array} $

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    $\begingroup$ Thanks for your answer. So, it makes sense to write $\displaystyle \sum\limits_{n=1}^\infty (-1)^n (\sum\limits_{k=n^2}^{(n+1)^2-1}\frac{1}{\sqrt{k}}-2)$ and check if this is a Leibniz series. $\endgroup$ – user90369 Apr 7 '17 at 15:00
  • $\begingroup$ Yes. Good idea. The Euler-Maclaurin sum formula would probably be useful. $\endgroup$ – marty cohen Apr 7 '17 at 17:52

It's alternating series (with periodically strange alternation, but alternation nonetheless) and thus converges according to alternating series test.

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    $\begingroup$ There are plenty of series with alternating signs which diverge. Take a modified harmonic series such that every 10th term is a negative addend. This series still diverges. $\endgroup$ – Just_a_fool Apr 7 '17 at 13:55
  • $\begingroup$ yes it converges $\endgroup$ – M. A. SARKAR Apr 7 '17 at 14:11
  • $\begingroup$ @mabmath. Could you elaborate, please ? $\endgroup$ – Claude Leibovici Apr 7 '17 at 14:14
  • $\begingroup$ To use alternating series test, we need more conditions but I think that series does not satisfy them $\endgroup$ – EUNSUNG LIM Apr 7 '17 at 14:17
  • $\begingroup$ yes it is , $ \sum_{n=1}^{\infty} \frac{(-1)^{[\sqrt{n}]}}{\sqrt n} $= $-1-\frac{1}{\sqrt 2} -\frac{1}{\sqrt 3}+\frac{1}{\sqrt 4}+\frac{1}{\sqrt 5}+\frac{1}{\sqrt 6}+\frac{1}{\sqrt 7}+\frac{1}{\sqrt 8} -\frac{1}{\sqrt 9}-... ........ $ . This is an alternating series and hence it is convegent. Because each successive term is numerically less than the preceding term and $ \lim_{n \rightarrow \infty } \frac{1}{\sqrt n} \rightarrow 0 $ $\endgroup$ – M. A. SARKAR Apr 7 '17 at 14:22

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