Is the series $\sum _{n=1}^{\infty } (-1)^n / {n^2}$ convergent or absolutely convergent? Is this series convergent or absolutely convergent? $$\sum _{n=1}^{\infty }\:(-1)^n \frac {1} {n^2}$$
Attempt:
I got this using Ratio Test: 
$$\lim_{n \to \infty} \frac{n^2}{(n+1)^2}$$
 A: 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. 
A: 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.
A: Hint: Use the integral test on $\sum_{k=1}^{\infty}\frac{1}{k^{2}}$.
Note that the ratio test is inconclusive in this case.
A: $\frac{1}{n^2}$ 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.
A: 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.  
