Calculating the sum of $\sum_{k=1}^{\infty}{\frac{(-1)^k}{k}}$ I am trying to find the sum of
$$\sum_{k=1}^{\infty}{\frac{(-1)^k}{k}}$$
I've proven that this converges using the Leibniz test, since
$a_n > 0$ and $\lim_{n\to\infty}{a_n} = 0$.
I am not sure how to go about summing this series up though. Every example I've seen up to now does some manipulation so that a geometric series pops out. I've been at it for a bit and I don't see how I could convert this to a geometric series to sum it up.

Using the Taylor series for the natural logarithm,
$$
\ln(x+1)=\sum_{n=1}^\infty\frac{(-1)^{n+1}x^n}{n}
$$
for $-1<x\le 1$. Abel's theorem guarantees that
$$
\ln2=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}.
$$
Hence,
$$\sum_{n=1}^\infty\frac{(-1)^{n}}{n}=-\ln2.
$$
