<s>[INST] Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k}  = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges, but what about the alternating harmonic series 
$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}  = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots \text{?} \tag{II}$$
Does it converge?   If so, what is its sum?
 [/INST] In this answer, I used only Bernoulli's inequality to show that
$$
\left(\frac{2n+1}{n+1}\right)^\frac{n}{n+1}
\le\left(1+\frac1n\right)^{n\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)}
\le\frac{2n+1}{n+1}\tag{1}
$$
The squeeze theorem and $(1)$, show that
$$
\exp\left[\lim\limits_{n\to\infty}\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)\right]=2\tag{2}
$$
That is,
$$
\begin{align}
\lim_{n\to\infty}\left(1-\frac12+\frac13-\frac14+\dots-\frac1{2n}\right)
&=\lim_{n\to\infty}\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)\\[6pt]
&=\log(2)\tag{3}
\end{align}
$$
</s>