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Is this series convergent or absolutely convergent? $$\sum _{n=1}^{\infty }\:(-1)^n \frac {1} {n^2}$$

Attempt:

I got this using Ratio Test:

$$\lim_{n \to \infty} \frac{n^2}{(n+1)^2}$$

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5 Answers 5

It is absolutely convergent since

$$ \Bigg| \frac{(-1)^n}{n^2} \Bigg| = \frac{1}{n^2}. $$

and $\sum_{n} \frac{1}{n^2}$ is a convergent series. Note that if the series converges absolutely then it converges.

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1  
@PedroTamaroff: You are right. Thanks for the comment. –  Mhenni Benghorbal Aug 13 at 20:25

Take the absolute value we get the convergent Riemann series $\sum_n \frac1{n^2}$. If you want a proof just notice that $$\frac1{n^2}\le \frac{1}{n(n-1)}=\frac{1}{n-1}-\frac1n,\;n\ge2$$ and the series $\sum_{n\ge2}\frac{1}{n-1}-\frac1n$ is convergent by telescopy.

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Hint: Use the integral test on $\sum_{k=1}^{\infty}\frac{1}{k^{2}}$.

Note that the ratio test is inconclusive in this case.

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$\frac{1}{n^2}$ converges by the p-series test. Since $\frac{1}{n^2}$ is always decreasing as $n$ increases the alternating series $(-1)^n\frac{1}{n^2}$ also converges. Hence, it is absolutely convergent.

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The root and ratio test, are testing of behavior similar to that of a geometric series. So for example $\sum \frac{n}{2^n}$ although not geometric is similar to a geometric and here root and ration will give an answer.

If the series does not behave like a geometric series the root and ratio are useless, second kind of series are those whose behavior is similar to a harmonic, like your example $\sum\frac{1}{n^2}$ here the integral, majorization/minorization, or the ratio comparison test will be successful.

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