How to show $\sum_{n=1}^\infty \frac{(-1)^{n}}{n^p},p \geq 1$ converges to a value $\geq -1$ I'm trying to find what $\sum_{n=1}^\infty \frac{(-1)^{n}}{n^p},p\geq1$ converges to. I know from the alternating series test that it converges. I've tried different values of $p$ and Mathematica and found that when p is very large, it converges to -1 and if $p$ equals 1, then the series converges to $-\ln(2)$. How can I find the convergence of this series?
 A: For $p=1$ you already got the correct answer. For $p>1$, the series is absolutely convergent, so we can write
$$
\sum_{n=1}^\infty \frac{(-1)^{n}}{n^p} = - \sum_{n=1}^\infty \frac{(-1)^{n+1}} {n^p} = - \left(\sum_{n=1}^\infty \frac{1}{n^p} - 2\sum_{k=1}^\infty \frac{1}{(2k)^p}\right),
$$
where first take out a $-1$ to have the series start with a positive term. Then we convert the sum with alternating signs into the same sum with positive signs, and then subtract double the value of the terms with the incorrect sign, which would be the numbers with even $n(=2k)$. 
The subtracted term can be rewritten to show that we only need to evaluate one infinite sum:
$$
\sum_{k=1}^\infty \frac{1}{(2k)^p} = \sum_{k=1}^\infty \frac{1}{2^p}\frac1{k^p} = \frac{1}{2^p} \sum_{k=1}^\infty \frac1{k^p} = \frac{1}{2^p} \sum_{n=1}^\infty \frac1{n^p}
$$
When we plug this into the previous equation, we get
$$
\sum_{n=1}^\infty \frac{(-1)^{n}}{n^p} = -\left( \sum_{n=1}^\infty \frac{1}{n^p} - 2\cdot\frac{1}{2^p} \sum_{n=1}^\infty \frac1{n^p}\right) = -\sum_{n=1}^\infty \frac1{n^p} \cdot (1-2^{1-p}) = (2^{1-p}-1)\sum_{n=1}^\infty \frac1{n^p}.
$$
On first sight, this looks like no help, as we have converted one infinite series into another. However $\sum_{n=1}^\infty \frac1{n^p}$ is the very important Rieman $\zeta$-Funktion, see https://en.wikipedia.org/wiki/Riemann_zeta_function. You can calculate its values with Mathematica and then apply the $(2^{1-p}-1)$ factor. 
