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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?

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  • $\begingroup$ $\lim_n a_n \neq 0$ $\endgroup$ – Mustafa Said Oct 30 '14 at 4:10
  • $\begingroup$ @Mary: Can you try to apply various tests (ratio and root tests, and the alternating series test ) to series $a_k$ defined in the second answer? $\endgroup$ – mike Oct 30 '14 at 6:16
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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.

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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.

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    $\begingroup$ The general term fo the series doesn't converge to zero, so what is your general observation saying? $\endgroup$ – Timbuc Oct 30 '14 at 4:55
  • $\begingroup$ @Timbuc: Please see my update. Thanks! $\endgroup$ – mike Oct 30 '14 at 5:02
  • $\begingroup$ 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? $\endgroup$ – Timbuc Oct 30 '14 at 5:05
  • $\begingroup$ 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. $\endgroup$ – mike Oct 30 '14 at 5:35
  • $\begingroup$ 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. $\endgroup$ – Timbuc Oct 30 '14 at 5:54

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