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Determine whether the series is conditionally convergent, absolutely convergent, or divergent.

$$\sum_{n=2}^{\infty}\frac{(-1)^n\sqrt{n}}{\ln(n)}$$

The absolute value of this sum is divergent by the divergence test, and it's inconclusive by the ratio and root tests, and the alternating series test doesn't apply because the sequence isn't decreasing. I think all of this only tells me that my conclusions are inconclusive. Is this correct? What can I do to actually determine if this series is absolutely convergent, conditionally convergent, or divergent?

Arnaud Mortier
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MaryG
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2 Answers2

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Note that $$ \lim_{n\to\infty}\frac{\sqrt{n}}{\ln(n)}\stackrel{L}{=}\lim_{n\to\infty}\frac{\frac{1}{2\sqrt{n}}}{\frac{1}{n}} =\frac{1}{2}\lim_{n\to\infty}\frac{n}{\sqrt{n}} =\frac{1}{2}\lim_{n\to\infty}\sqrt{n}=\infty $$ This implies that your sum diverges.

Brian Fitzpatrick
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This is not a complete solution, just an observation.

We can rewrite the series as:

$$\sum_{n=2}^{\infty}b_n:=\sum_{n=2}^{\infty}\frac{(-1)^n\sqrt{n}}{\ln(n)}=\sum_{k=1}^{\infty}\left(\frac{\sqrt{2k}}{\ln(2k)}-\frac{\sqrt{2k+1}}{\ln(2k+1)}\right):=\sum_{k=1}^{\infty}a_k$$

It is shown in the first answer that general term $b_n$ is going to $\infty$ when $n\to\infty$: $$\lim_{n\to\infty}b_n\to \infty.$$

But we should not jump to conclusion based on this fact.

Here we can show that the general term $a_k$ of the transformed series is going to 0 when $k\to\infty$:

$$\lim_{k\to\infty}a_k=\frac{2+\log 1-\log{(2k)}}{2\sqrt{2k}(\log 1-\log{(2k)})^2}\to0.$$

Thus for alternating series $\sum_{n=2}^{\infty}(-1)^n b_n$ it is better to look at the transformed one: $\sum_{k=1}^{\infty}(b_{2k}-b_{2k+1}):=\sum_{k=1}^{\infty}a_k$.

EDIT: Here is an example that behaves similar to the series in OP.

$$S=\sum_{n=2}^{\infty}b_n=\sum_{n=2}^{\infty}\left((-1)^{n+1}[n/2]+\frac{1}{n^2}\right)\tag{5}$$

where $[n/2]$ is the largest integer that is smaller or equal to $n/2$.

Thus $[(2k)/2]=k$ and $[(2k+1)/2]=k$, for $k\in \mathbb{N}$.

We notice that $|b_n|\to [n/2] \to\infty$ when $n\to\infty$.

On the other hand we have $$b_{2k}+b_{2k+1}=\left((-1)^{2k+1}[(2k)/2]+\frac{1}{(2k)^2}\right)+\left((-1)^{2k+2}[(2k+1)/2]+\frac{1}{(2k+1)^2}\right)$$ $$=\left(-k+\frac{1}{(2k)^2}\right)+\left(k+\frac{1}{(2k+1)^2}\right)=\frac{1}{(2k)^2}+\frac{1}{(2k+1)^2}:=a_k$$

Thus we may rewrite $S$ by grouping $b_{2k}$ and $b_{2k+1}$ as:

$$S=\sum_{k=1}^{\infty}a_k=\sum_{k=1}^{\infty}\left(\frac{1}{(2k)^2}+\frac{1}{(2k+1)^2}\right)\tag{6}$$

We also notice that $a_k\to 0$ when $k\to\infty$.

It is obvious that series $S$ is convergent. Thus using the asymptotic property of the general term $b_n$ alone to decide the convergence is erroneous.

mike
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    The general term fo the series doesn't converge to zero, so what is your general observation saying? – Timbuc Oct 30 '14 at 04:55
  • @Timbuc: Please see my update. Thanks! – mike Oct 30 '14 at 05:02
  • So your answer contains two contradicting statements, @Mike: the same series is written in two different but equivalent (otherwise the whole thing is wrong) ways, yet in one the general term goes to zero whether in the other it doesn't? – Timbuc Oct 30 '14 at 05:05
  • What I am trying to say is that series $a_k$ represents the original series in a better way. Conclusion made based on the behavior of the general term $b_n$ is less reliable. – mike Oct 30 '14 at 05:35
  • My doubt remains, @mike: we have the same thing represented in two way that behave differently: one converges to zero, the other one diverges to infinity. I don't think that's logical. I haven't checked your work, but I think the above isn't possible. – Timbuc Oct 30 '14 at 05:54
  • @Timbuc: If you numerically plot the truncated series $B(N)=\sum_{n=2}^{N}b_n$ and $A(K)=\sum_{k=1}^{K}b_k$ with $N=2K=200,300,400,..$, you will see what I am talking about. – mike Oct 30 '14 at 05:59
  • again: how can you explain that you write the same series in two different ways, and in one the general term sequence converges to zero but in the other the general term sequence diverges to infinity?? Unless I'm misunderstanding badly something, this can't be. Please do address this doubt. – Timbuc Oct 30 '14 at 10:40
  • Can someone address Timbuc's doubt? I feel like that I can not properly answer Timbuc's question. @Timbuc: can you ask the author of the first answer? – mike Oct 30 '14 at 11:29
  • it is you that wrote this! Either you made a mistake when writing the third sum in your first line of calculations, or else it isn't true that the limit if this expression is infinite. You must prove this, not anyone else, as it is you who's claiming that. I can easily prove the limit of the original general term sequence is zero, so either that third expression is wrong or else its limit is again zero. You prove your point, don't ask anyone else to do it. – Timbuc Oct 30 '14 at 11:32
  • I added one example in my answer. This example also showed you the dual behavior of a series. Please let me know if there is anything that is not correct. – mike Oct 30 '14 at 16:34
  • after reviewing your answer, I think I know where the mistake is: the second equality in your first line of mathematics is false: you are putting parentheses to a series that converges conditionally! That's forbidden, of course. This is what I meant in my first comment: ifyou write equality between two series whose general term is expressed differently, it cannot be possible that in one case the general term goes to infinity and in the other one it doesn't. – Timbuc Oct 30 '14 at 16:54
  • OK. I got it. Thanks! – mike Oct 30 '14 at 17:01
  • "But we should not jump to conclusion based on this fact." Why not? – Arnaud Mortier Jun 14 '18 at 23:39