<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] Here is another proof, based on the formula
$$\frac{1}{1+x}=\frac{(-1)^nx^n}{1+x}+\sum_{k=0}^{n-1}(-1)^kx^k.$$
Integrating both sides over $[0,t]$ gives
$$\ln(1+t)=\int_0^t\frac{(-1)^nx^n}{1+x}\,dx+\sum_{k=1}^n(-1)^{k+1}\frac{t^k}{k}.$$
Setting $t=1$ shows that the partial sums $s_n$ of the alternating harmonic series are given by
$$s_n=\ln2+(-1)^n\int_0^1\frac{x^n}{1+x}\,dx.$$
But on $[0,1]$, we have $0\leq x^n(1+x)^{-1}\leq x^{n}$, so
$$0\leq\int_0^1\frac{x^n}{1+x}\,dx\leq\int_0^1x^{n}\,dx=\frac{1}{n+1}.$$
Hence $$\ln2-\frac{1}{n+1}\leq s_n\leq\ln2+\frac{1}{n+1}.$$ So $s_n\to\ln 2$ as $n\to\infty$.
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