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I'm looking for help checking if the series:

$$\sum_{n=2}^{\infty}(-1)^nn\ln^2\left(\frac{n+1}{n-1}\right)$$

converges absolutely, conditionally or diverges.

Here are my thoughts:

For conditional convergence I should use Leibnitz theorem.

I can show that $\lim_\limits{{n\to \infty}} a_n =0$ but when I tried to show it is monotonous I hit a dead end (I tried by finding the derivative or looking at $a_n- a_{n+1}$).

For absolute convergence I tried to use different tests but got limit of 1 so I couldn't learn from that.

Any ideas?

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    $\begingroup$ For absolute convergence, write $\frac{n+1}{n-1}=1+\frac{2}{n-1}$, and use the fact that $\log(1+x)\approx x$ near zero to conclude that the absolute value of the general term is of the order of magnitude of $\frac{1}{n}$. $\endgroup$ – uniquesolution Aug 5 '17 at 17:06
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$$\log\frac{n+1}{n-1} = \log\frac{1+\frac{1}{n}}{1-\frac{1}{n}} = 2\,\text{arctanh}\frac{1}{n} = \frac{2}{n}+O\left(\frac{1}{n^3}\right) \tag{1}$$ leads to $$ n\log^2\frac{n+1}{n-1} = \frac{4}{n}+O\left(\frac{1}{n^3}\right) \tag{2} $$ so the given series is conditionally convergent but not absolutely convergent, i.e. has exactly the same convergence behaviour of $4\sum_{n\geq 2}\frac{(-1)^n}{n}$, since $\sum_{n\geq 2}\frac{1}{n^3}$ is clearly absolutely convergent.

Are you interested in a exact evaluation, too?

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  • $\begingroup$ that's great as it is. thank you $\endgroup$ – segevp Aug 5 '17 at 17:13
  • $\begingroup$ @Jack D'Aurizio...is your site in English or in (I pressume from your name)Italian? $\endgroup$ – Marios Gretsas Aug 5 '17 at 17:16
  • $\begingroup$ @MariosGretsas: In Italian, but I should create an English version, too. The last version of my course notes is in English, at the very least. $\endgroup$ – Jack D'Aurizio Aug 5 '17 at 17:17
  • $\begingroup$ in which courses are your notes? $\endgroup$ – Marios Gretsas Aug 5 '17 at 17:19
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    $\begingroup$ @MariosGretsas: drive.google.com/file/d/0BxKdOVsjsuEwdjBEM1dpRkhMa2s/view are related with an extra-curricular course for students of Mathematics (1st and 2nd year) at the University of Pisa; I kept it in 2016 and I will keep it this year, too, from October. $\endgroup$ – Jack D'Aurizio Aug 5 '17 at 17:22
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$\ln \left(\frac{n+1}{n-1}\right) = \frac{2}{n-1} + O\left(\frac{1}{n^2}\right)$. Can you conclude from that ?

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$$\ln (n+1)=\ln (n)+\ln (1+1/n) $$ $$=\ln (n)+\frac {1}{n}-\frac {1}{2n^2}+\frac {1}{3n^3}(1+\epsilon (n)) $$ $$\ln (n-1)=\ln (n)-\frac {1}{n}-\frac {1}{2n^2}-\frac {1}{3n^3}(1+\epsilon (n)) $$

thus

$$\ln (\frac {n+1}{n-1})=\frac {2}{n}+\frac {2}{n^3}(1+\epsilon (n)) $$ and $$\ln^2 (\frac {n+1}{n-1})=\frac {4}{n^2}+\frac {8}{n^4}(1+\epsilon (n)) $$

hence $$u_n=\frac {(-1)^n4}{n}+\frac {(-1)^n8}{n^3}(1+\epsilon (n)) $$

the sum of a conditionally convergent $( \sum(-1)^n\frac {4}{n} ) \;$ and an absolutely convergent $( \sim \frac {8}{n^3}) \;$ is a conditionally convergent series.

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