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I came across the following series in my math homework (Fourier Series):

Does the following series converge or diverge? If converges, does it converge absolutely?

$\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{n^2+3}$

Typically, I would be well equipped to answer the question, however the "n=$-\infty$" is giving me trouble. Normally, if "n=$0$", the alternating series test could show convergence, and a direct comparison test with a p-series could show absolute convergence. How does the "$-\infty$" change the problem, if at all?

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Since the summand $f(n)$ satisfies $$ f(-n)=-f(n) $$ one may just study the convergence over $n \in [0,\infty)$.

Then the series is absolutely convergent by the $p$-test and the given series is convergent.

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\begin{equation} \sum\limits_{n= -\infty}^0 \frac{(-1)^n}{n^2 + 3} = \sum\limits_{n= 0}^{+\infty} \frac{(-1)^n}{n^2 + 3} \end{equation} So \begin{equation} \sum\limits_{n= -\infty}^{+\infty} \frac{(-1)^n}{n^2 + 3} = \underbrace{ \sum\limits_{n= -\infty}^0 \frac{(-1)^n}{n^2 + 3} }_{n = -\infty \ldots 0} + \underbrace{ \sum\limits_{n= 0}^{+\infty} \frac{(-1)^n}{n^2 + 3} }_{n = 0 \ldots +\infty} - \underbrace{ \frac{(-1)^0}{0^2 + 3} }_{n=0} = 2\sum\limits_{n= 0}^{+\infty} \frac{(-1)^n}{n^2 + 3} - \frac{1}{3} \end{equation} The series \begin{equation} \sum\limits_{n= 0}^{+\infty} \frac{(-1)^n}{n^2 + 3} \end{equation} is absolutely convergent due to the $p-$test. So your series converges.

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