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?

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

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