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?
Answer
There are actually two “more direct” proofs of the fact that this limit is $\ln (2)$.
First Proof Using the well knows (typical induction problem) equality:
$$\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+…+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n} \,.$$
The right side is $\frac{1}{n} \left[ \frac{1}{1+\frac{1}{n}}+ \frac{1}{1+\frac{2}{n}}+..+\frac{1}{1+\frac{n}{n}} \right]$ which is the standard Riemann sum associated to $\int_0^1 \frac{1}{1+x} dx \,.$
Second Proof Using $\lim_n \frac{1}{1}+\frac{1}{2}+…+\frac{1}{n}-\ln (n) =\gamma$.
Then
$$\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+…+\frac{1}{2n-1}-\frac{1}{2n}=
\left[ \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+…+\frac{1}{2n-1}+\frac{1}{2n} \right]-2 \left[\frac{1}{2}+\frac{1}{4}…+\frac{1}{2n} \right] $$
$$= \left[ \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+…+\frac{1}{2n-1}+\frac{1}{2n} \right]-\ln(2n) – \left[\frac{1}{1}+\frac{1}{2}…+\frac{1}{n} \right]+\ln(n) + \ln 2 \,.$$
Taking the limit we get $\gamma-\gamma+\ln(2)$.
Attribution
Source : Link , Question Author : Isaac , Answer Author : Pete L. Clark