Absolute Convergence of Homework Series: Real Parameter p

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Homework Statement


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

p is a real parameter, determine when the series converges absolutely/non-absolutely

The Attempt at a Solution



I tried to do the limit \lim_{n\rightarrow \infty} \frac{\ln \left(1+\frac{(-1)^n}{n^p}\right)}{\frac{(-1)^n}{n^p}}, which is equal to one and this suggests that the series coverges if p is positive (limit comparison test). But then I'm not sure how to determine the absolute/non-absolute convergence. Could you help me please? Thanks very much in advance!
 
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Try to split the series into two parts - the even numbers and the odd ones and see if they both converge.
 
i made a mistake in the first post, the limit comparison test is applicable only to non-negative series. then the limit should be like:
\lim_{n\rightarrow \infty} \frac{\left|\ln \left(1+\frac{(-1)^n}{n^p}\right)\right|}{\frac{1}{n^p}}

which is equal to zero when p is positive. However, the lower series converges when p>1. Therefore, the original logarithm series converges absolutely for p>1.

The textbook then says that the series converges non-absolutely also for 1/2 < p <= 1. But I can't prove it - do you have any ideas?
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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