Let $\alpha \in \mathbb{R}$. Examine the (absolute) convergence of the following series, depending on $\alpha$

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

My attempt:

Let's start with absolute convergence. For $\alpha > -1$, we have:

$$\frac{1}{n^{\alpha} (n^2 +1)} \le \frac{1}{n^{\alpha +2}}$$

and the result follows by comparing to the convergent series $\sum\frac{1}{n^{\alpha +2}}$

For $\alpha \leq -1$, there is divergence:

$$\frac{1}{n^{\alpha}(n^2 +1)} = \frac{1}{n^{\alpha+2} + n^{\alpha}} \geq \frac{1}{n^{\alpha+2} + n^{\alpha+2}} = \frac{1}{2n^{\alpha+2}}$$

and by comparing to the divergent series $\sum \frac{1}{2n^{\alpha+2}}$ the result follows.

So, we also have convergence for the given series whenever $\alpha > -1$

Whenever $\alpha \in (-2,-1]$, the sequence $\frac{1}{n^\alpha(n²+1)} \to 0$ and decreases (this can be seen using derivatives). Hence, the series surrenders for the alternating series test.

For $\alpha \leq -2$,we have: $\frac{(-1)^n}{n^{\alpha}(n^2+1)} \not\to 0$. Hence, divergence.

Is this correct?

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    $\begingroup$ Didn't quite understand what do you mean for "surrenders the test"... XD $\endgroup$ – Macrophage Dec 30 '17 at 11:54
  • $\begingroup$ It means we can deduce convergence from that test :P $\endgroup$ – user370967 Dec 30 '17 at 11:57
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    $\begingroup$ Then, the proof looks fine! $\endgroup$ – Macrophage Dec 30 '17 at 12:01
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    $\begingroup$ @Math_QED when you have negative exponent I tink that it's often convenient to change the varible in order to deal with positive numbers. $\endgroup$ – gimusi Dec 30 '17 at 12:04
  • $\begingroup$ Yes, I saw that little trick in your answer. Will keep it in mind for future use :) $\endgroup$ – user370967 Dec 30 '17 at 12:05

I agree with your solution.

For $\alpha \geq 0$ the serie clearly converges absolutely.

For $\alpha < 0$ let $\beta = -\alpha$

$$\frac{(-1)^n}{n^\alpha (n^2+1)}=\frac{n^{\beta}(-1)^n}{n^2+1}$$

which converges for $0<\beta<2$ and diverges otherwise.


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