Does This Alternating Series Diverge?

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Homework Statement


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

Homework Equations


This is in the section covering alternating sequences. Leibniz's rule, conditional/absolute convergence, Dirichlet's test, and Abel's tests were all covered.

The Attempt at a Solution



I don't know what to apply here, it seems like none of the tests I have learned are applicable. Most of the theorems in this section covered sufficient conditions for convergence, which I can't turn back on themselves RAA to prove that this diverges. Obviously the series does not absolutely converge, and the even terms are greater in absolute value than the odd terms, so it will keep increasing, however I think there's a problem with approaching it this way since it is essentially grouping terms of the infinite sequence within parenthesis, which is not strictly allowed.
 
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I think you can try two ways:
1) since you are interested in its behaviour when n is great, you can find an asymptotic approximation and then analyse it with Leibniz's rule;
or
2) you can find a series \sum b_n such that b_n\geq a_n (work on the denominator, trying to eliminate (-1)^n )
 
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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