Convergence of series whose general term is $u_n = \sum_{k=n}^{\infty}\frac{\left(-1\right)^k}{k^2}$ Let $\left(u_n\right)$ the sequence defined for all $n \geq 1$ by:
$$u_n = \sum_{k=n}^{+\infty}\frac{\left(-1\right)^k}{k^2}$$
I've shown that $u_n$ has a sense and that $\sum_{n =1}^\infty u_n$ converges. I'm asked to compute $\sum_{n=1}^{\infty}u_n$. What I did is to consider the partial sum $S_n$
$$
S_n = \sum_{k=1}^{n}u_k = \sum_{k=1}^{\infty}\frac{\left(-1\right)^k}{k^2} + \sum_{k=2}^{\infty}\frac{\left(-1\right)^k}{k^2} + \dots + \sum_{k=n}^{\infty}\frac{\left(-1\right)^k}{k^2}
$$
I dont really know how to proceed further. Any help ?
 A: Note the similarities between the $u_n$. The partial sums “stack” up. For instance:

$$\begin{align}u_1+u_2&=\sum_{k=1}^\infty\frac{(-1)^k}{k^2}+\sum_{k=2}^\infty\frac{(-1)^k}{k^2}\\&=\frac{(-1)^1}{1^2}+\sum_{k=2}^\infty\frac{(-1)^k}{k^2}+\sum_{k=2}^\infty\frac{(-1)^k}{k^2}\\&=-1+2\sum_{k=2}^\infty\frac{(-1)^k}{k^2}\end{align}$$

In general, in a partial sum $S_n$, all the summands with $k\ge n$ will be counted $n$ times and the summands with $k<n$ are counted $k$ times, causing a cancellation $k/k^2=k$.
$$\begin{align}S_n&=\sum_{m=1}^n u_m\\&=n\sum_{k=n}^\infty+(n-1)\frac{(-1)^{n-1}}{(n-1)^2}+\cdots\\&=\sum_{k=1}^{n-1}\frac{(-1)^k}{k}+n\sum_{k=n}^\infty\frac{(-1)^k}{k^2}\end{align}$$
We know that $\sum_{k=1}^\infty\frac{(-1)^k}{k}$ converges to $-\ln2$, and by the alternating series test’s proof: $$\left|n\sum_{k=n}^\infty\frac{(-1)^k}{k^2}\right|=\left|\sum_{k=n}^\infty\frac{(-1)^k}{k\cdot(k/n)}\right|\le\frac{1}{n}$$
Therefore:

$$\begin{align}|S_n+\ln 2|&\le\frac{1}{n}+\left|\ln2+\sum_{k=1}^{n-1}\frac{(-1)^k}{k}\right|\\&\overset{n\to\infty}{\longrightarrow}0\end{align}$$

Giving a proof of convergence and an evaluation (as $-\ln2$) in one step.
A: Let $S_k=\sum_{i=1}^k \frac{(-1)^i}{i^2},~~S=\sum_{i=1}^\infty \frac{(-1)^i}{i^2}=\sum_{i=1}^\infty (-1)^i a_i$, so $u_k=S-S_{k-1}$, since $(-1)^ia_i$ is alternating, we have
$$|u_k|=|S-S_{k-1}|\le a_k=\frac{1}{k^2}$$
so $$\sum|u_k|\le \sum \frac{1}{k^2}.$$
So the series $\sum u_k$ is absolutely convergent.
