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The task is to check for which $\alpha$ does the series converge.

A) $\sum_{n=1}^\infty \frac{(-1)^n}{n^\alpha}$

B) $\sum_{n=1}^\infty \frac{(-1)^n}{n+\alpha}$

C) $\sum_{n=1}^\infty \frac{(-1)^n * (log_en)^\alpha}{n}$

So with first one I went with dividing the problem to cases.

  1. $\alpha=1$

$\lim\limits_{n \to \infty}\frac{(-1)^n}{n}=0$ by Alternating series test it converges.

  1. $\alpha>1$

$\lim\limits_{n \to \infty}\frac{(-1)^n}{n^\alpha}=0$

and now with Ratio test

$\lim\limits_{n \to \infty}\frac{(-1)^{n+1}}{(n+1)^\alpha}*\frac{n^\alpha}{(-1)^n}=\lim\limits_{n \to \infty}-(\frac{n}{(n+1)})^\alpha=-1$

so it meets the cryterion, and converges.

  1. $1>\alpha>0$

Analogously to point 2. it converges.

  1. $\alpha<0$ Here I take

$\alpha=-\beta$

$\beta>0$

And at first I try to check if the series satisfy Limit of the summand test and...i got stuck with

$\lim\limits_{n \to \infty}(-1)^nn^\beta$

By Wolfram Alpha I know that it converges to infinity, but I do not know how to prove that... and how to approach next two series. I would be grateful for any hints how to deal with that task :)

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    $\begingroup$ You use HIndu-Arabic numbers both for the questions and for the cases of question 1. It would be less confusing if you used letters for the cases, such as a,b,c. And for case 2 (or b), the limit of the absolute value of the ratio is $1$, which makes the test inconclusive.. You need another approach to solve that case. $\endgroup$ – Rory Daulton Nov 21 '15 at 13:02
  • $\begingroup$ I am a bit unsure, about that 1. If $(-1)^{n+1} = (-1)^n*(-1)$, and after It as I divide it by $(-1)^n$ I have bare $-1$ which I cen take before the whole limit. So at the end the limit is not 1, but -1. Maybe I make somewhere huge mistake, but I don't see that. $\endgroup$ – Kiwi Nov 21 '15 at 13:07
  • $\begingroup$ The ratio test uses the absolute values of the ratios, not the ratios themselves, as you can see in my previous link. Anyway, if the limit of the ratios themselves is $-1$ we have insufficient information to decide convergence. All the series $\sum\frac{(-1)^n}n$, $\sum(-1)^n$, and $\sum(-1)^nn$ have the limiting ratio $-1$, but the first converges, the second alternates and is bound, and the third alternates and is unbound. You need more information. $\endgroup$ – Rory Daulton Nov 21 '15 at 13:10
  • $\begingroup$ Ok, my mistake. I got to check that one again. Thanks! $\endgroup$ – Kiwi Nov 21 '15 at 13:13
  • $\begingroup$ Raabe–Duhamel's test seems solving the problem. $\endgroup$ – Kiwi Nov 21 '15 at 13:22
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For the first question: if $\alpha < 0$, then $$ \frac{(-1)^{n}}{n^{\alpha}} = (-1)^{n}n^{|\alpha|}; $$ given any $M > 0$, we have $|(-1)^{n}n^{|\alpha|}| = n^{|\alpha|} > M$ if $n > M/\log |\alpha|$.

For the second question: if $\alpha \in \Bbb{R}$, then there is some $N \geq 1$ such that the sequence $( \frac{1}{n+\alpha} )_{n \geq N}$ is decreasing and $\to 0$ as $n \to \infty$; hence by Leibniz's test the corresponding series converges.

For the third question: by proof for all $a,b > 0$ we have $(\log x)^{a}/x^{b} \to 0$ as $x \to \infty$; hence for every $\alpha \in \Bbb{R}$ there is some $N \geq 1$ such that the sequence $((\log n)^{\alpha}/n)_{n \geq N}$ is decreasing and $\to 0$ as $n \to \infty$, implying by Leibniz's test that the corresponding series converges.

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