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Let $0 < \alpha <1$ and consider the summation

$$\displaystyle\sum_{n = 1}^\infty \ln \bigg|\frac{\alpha - n}{\alpha + n}\bigg|.$$

Am I correct in thinking that the individual terms in this series go towards zero as $n$ gets large because the argument supplied to $\ln$ goes towards $1$? What does the summation converge to?

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    For fixed $\alpha$, the summands are $\sim \frac1{n 2}$, hence the series does converge. – 2017-02-13
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    @HagenvonEitzen You mean that the terms are more like $1/n$ and the series diverge, right? (also, to OP, you mean $0<\alpha<1$, right?) – 2017-02-13
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    $$\lim_{\infty}\ln \bigg|\frac{\alpha - n}{\alpha + n}\bigg|=0$$ is a necessary condition for convergentness not sufficient. – 2017-02-13
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    @MyGlasses Ah, sign error! I thought the odd powers would cancel, but the even ones do ... – 2017-02-13

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For $0<\alpha<1$ and $n\geq 1$ you can write $$ \ln\Bigl|\frac{\alpha-n}{\alpha+n}\Bigr|=\ln\frac{n-\alpha}{n+\alpha}=\ln\Bigl(1-\frac{2\alpha}{n+\alpha}\Bigr). $$ Since $$ \lim_{n\to+\infty} \frac{\ln\Bigl(1-\frac{2\alpha}{n+\alpha}\Bigr)}{-\frac{2\alpha}{n+\alpha}}=1 $$ and all terms in your series are negative we conclude that your series converges if and only if the series $$ \sum_{n=1}^{+\infty}\frac{2\alpha}{n+\alpha} $$ converges, which we(?) know that it does not.

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    Thanks I see how to approach it now! – 2017-02-13
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$\displaystyle\sum_{n = 1}^\infty \ln \bigg|\frac{\alpha - n}{\alpha + n}\bigg|. $

For any real $a$, once $n > |a|$,

$\begin{array}\\ \ln \bigg|\frac{a-n}{a+n}\bigg| &=\ln \frac{n-a}{n+a}\\ &=\ln \frac{1-a/n}{1+a/n}\\ &=\ln (1-a/n)-\ln(1+a/n)\\ &=-\frac{a}{n}-\frac{a^2}{2n^2}+O(\frac{a^3}{n^3})-(\frac{a}{n}-\frac{a^2}{2n^2}+O(\frac{a^3}{n^3}))\\ &=-\frac{2a}{n}+O(\frac{a^3}{n^3})\\ \end{array} $

and the sum of these diverges.