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?

Here is one proof, using how ${ \log (1+x) }$ is "sandwiched" between its Taylor polynomials on ${ [0,1] }.$
Consider ${ \log(1+x) }$ on ${ (-1, \infty) }.$ Its Taylor polynomials about $0$ are ${ S _n (x) = x - \frac{x ^2}{2} + \ldots + (-1) ^{n+1} \frac{x ^n}{n}. }$
Notice errors ${ \varepsilon _n (x) := \log(1+x) - \left( x - \frac{x ^2}{2} + \ldots + (-1) ^{n+1} \frac{x ^n}{n} \right) }$ satisfy ${ \varepsilon _n (0) = 0, }$ and ${ \varepsilon _n ' (x) }$ ${ = \frac{1}{1+x} -(1-x + \ldots + (-1) ^{n+1} x ^{n-1}) }$ ${ = \frac{1 - (1+ (-1) ^{n+1} x ^n)}{1+x} }$ ${ = \frac{ (-1) ^n x ^n}{1+x} }$ is $\geq 0$ for ${ \lbrace  x \in [0, \infty); n \text{ even} \rbrace }$ and $\leq 0$ for ${ \lbrace x \in [0, \infty); n \text{ odd} \rbrace }.$
So ${ \varepsilon _{n} (x) }$ is ${ \geq 0 }$ for ${ \lbrace x \in [0, \infty); n \text{ even} \rbrace }$ and ${ \leq 0 }$ for ${ \lbrace x \in [0, \infty); n \text{ odd} \rbrace }.$
So ${ S _{2n-1} (x) \geq \log(1+x) \geq S _{2n} (x) }$ for ${ x \in [0, \infty), n \in \mathbb{Z} _{\gt 0} }.$
Further ${ S _{2(n+1) -1} (x) - S _{2n-1} (x) }$ ${ = - \frac{x ^{2n}}{2n} + \frac{x ^{2n+1}}{2n+1} }$ ${ = x ^{2n} ( \frac{x}{2n+1} - \frac{1}{2n}) \leq 0}$ if ${ \underline{ x \in [0,1] } },$ ie sequence ${ (S _{2n-1} (x)) _{n \geq 1} }$ is decreasing if ${ x \in [0,1] }.$
Similarly ${ S _{2(n+1)} (x) - S _{2n} (x) }$ ${ = \frac{x ^{2n+1}}{2n+1} - \frac{x ^{2n+2}}{2n+2} }$ ${ = x ^{2n+1} (\frac{1}{2n+1} - \frac{x}{2n+2}) \geq 0 }$ if ${ \underline{ x \in [0,1] } },$ ie sequence ${ (S _{2n} (x) ) _{n \geq 1} }$ is increasing if ${ x \in [0,1] }.$
Fix ${ x \in [0,1] }.$
Now sequence ${ (S _{2n} (x) ) _{n \geq 1} }$ is increasing and ${ (S _{2n-1} (x) ) _{n \geq 1} }$ decreasing, with ${ S _{2n-1} (x) \geq \log(1+x) \geq S _{2n} (x) }.$
So say ${ \lim _{n \to \infty} S _{2n} (x) = \ell _e }$ and ${ \lim _{n \to \infty} S _{2n-1} (x) = \ell _o }.$ But ${ S _{2n-1} (x) - S _{2n} (x) = \frac{x ^{2n}}{2n} \leq \frac{1}{2n} \to 0 }$ as ${ n \to \infty }.$ So ${ \ell _o - \ell _e = 0 },$ ie ${ \ell _o = \ell _e }.$
This gives ${ \ell _o = \ell _e = \log(1+x), }$ making ${ \log(1+x) = \lim _{j \to \infty} S _j (x). }$

Especially ${ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} +  \ldots }$ converges, and to ${ \log(2) }.$

