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.

Since an alternating sum is the difference of the non-alternating sum and twice the sum of the even terms, we get
$$
\begin{align}
\sum_{k=1}^{2n}\frac{(-1)^{k-1}}k
&=\sum_{k=1}^{2n}\frac1k-2\sum_{k=1}^n\frac1{2k}\\
&=\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac1k\\
&=\sum_{k=n+1}^{2n}\frac1k\tag{1}
\end{align}
$$
By comparison to an integral, we have
$$
\log\left(\frac{2n+1}{n+1}\right)=\int_{n+1}^{2n+1}\frac1x\,\mathrm{d}x\le\sum_{k=n+1}^{2n}\frac1k\le\int_n^{2n}\frac1x\,\mathrm{d}x=\log(2)\tag{2}
$$
Apply the Squeeze Theorem to $(2)$ and incorporate $(1)$ and we get
$$
\sum_{k=1}^\infty\frac{(-1)^{k-1}}k=\log(2)\tag{3}
$$
