$\sum^{\infty}_{n=1} \frac{(-1)^n}{n^2}$ prove convergent without alternating test $$\sum^{\infty}_{n=1} \frac{(-1)^n}{n^2}$$
I know how to prove this one is convergent by alternating test. like Let $b_n = \frac{1}{n^2}$ and limit of this sum is $0$ and also decreasing. 
However, my professor doesn't allow me to use that I did not learn yet. So I cannot use alternating test(I can use p-series, comparison, ratio test). I don't know how to solve this one..... without alternating test. 
 A: You know about $p$-series, so you know the series $\sum_{n=1}^\infty \frac{1}{n^2}$ converges.
Proving convergence of your alternating series $\sum_{n=1}^\infty \frac{(-1)^n}{n^2}$ is equivalent to proving the sequence of its partial sums, $(S_n)_{n\geq 1}$ is Cauchy.
It will be helpful to compare it to the $p$-series you do know something about; so observe that for any $m\geq 1$ ,$p\geq 0$,
$$
\left\lvert\sum_{n=m+1}^{m+p} \frac{(-1)^n}{n^2} \right\rvert
\leq \sum_{n=m+1}^{m+p} \left\lvert\frac{(-1)^n}{n^2} \right\rvert
= \sum_{n=m+1}^{m+p} \frac{1}{n^2} \tag{1}
$$
by the triangle inequality.
So fix $\varepsilon > 0$. Since $\sum_{n=1}^\infty \frac{1}{n^2}$ converges, it is Cauchy, so there exists $m\geq 1$ such that, for all $p\geq 0$, $\sum_{n=m+1}^{m+p} \frac{1}{n^2} \leq \varepsilon$. By the above (1), for the same $m$ and any $p\geq 0$, we have 
$$
\lvert S_{m+p}-S_m\rvert = \left\lvert\sum_{n=m+1}^{m+p} \frac{(-1)^n}{n^2} \right\rvert
\leq \varepsilon.
$$
This shows $(S_n)_{n\geq 1}$ is Cauchy, thus convergent.
A: Note that we can write
$$\begin{align}
\sum_{n=1}^{2N} \frac{(-1)^n}{n^2}&=\sum_{n=1}^{2N} \frac{1+(-1)^n}{n^2}-\sum_{n=1}^{2N} \frac{1}{n^2}\\\\
&=\sum_{n=1}^N\frac{2}{(2n)^2}-\sum_{n=1}^{2N} \frac{1}{n^2}\\\\
&=\frac12 \sum_{n=1}^N\frac{1}{n^2}-\sum_{n=1}^{2N} \frac{1}{n^2}\tag1
\end{align}$$
Since we know that both of the partial sums on the right-hand side of $(1)$ converge, then the partial sum on the left-hand side must converge also. In fact, we have from $(1)$
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}=-\frac12 \sum_{n=1}^\infty \frac{1}{n^2}$$
And that completes the proof.

We have a much more general way to approach this problem.  We will now show that if a series converges absolutely, then it converges.  
To show this, we assume that $\sum_{n=1}^\infty |a_n|$ converges.  Now, since $\sum_{n=1}^\infty |a_n|$ converges, then $\sum_{n=1}^\infty 2|a_n|$ converges also.
Note that we have the inequalities $0\le a_n+|a_n|\le 2|a_n|$.  Since the sequence $a_n+|a_n|$ is non-negative, then 
$$0\le \sum_{n=1}^N(a_n+|a_n|)\le \sum_{n=1}^N2|a_n| \tag 2$$
The sequence of partial sums $S_N=\sum_{n=1}^N(a_n+|a_n|)$ in $(2)$ is monotonically increasing and bounded above by $\sum_{n=1}^\infty 2|a_n|$.  Therefore, $S_N$ converges.
Finally, we can write
$$\sum_{n=1}^N a_n=\sum_{n=1}^N (a_n+|a_n|)-\sum_{n=1}^N |a_n| \tag 3$$
Inasmuch as the sequence of partial sums on the right-hand side of $(3)$ converge, the sequence of partial sums on the left-hand side of $(3)$ converge also.
We have just shown that any absolutely convergent sequence also converges.
A: $$\sum \frac{(-1)^n}{n^2} = -1 + \frac{1}{4} - \frac{1}{9} + \frac{1}{16} \cdots = \sum \frac{1}{(2n)^2}- \frac{1}{(2n-1)^2} = \frac{1}{4}\sum\frac{1}{n^2} - \sum \frac{1}{(2n-1)^2}$$
The latter two sums are convergent by $p$ series test.  
