How does this series converge to log(2)?

Question

I know that the series

$$\frac1{1*2}+\frac1{3*4}+\frac1{5*6}+\frac1{7*8}+\dots$$

converges to $\log(2).$

But how can I prove it?

Answer 1

For $-1<x\le1,$ $$\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots$$

Set $x=1$

Here $r$th term $$=\frac1{(2r-1)(2r)}=\frac{2r-(2r-1)}{(2r-1)(2r)}=\frac1{2r-1}-\frac1{2r}$$

See also : Convergence for log 2

Answer 2

Another approach. Since: $$\frac{1}{2n(2n-1)}=\frac{1}{2\binom{2n}{2}}=\frac{\Gamma(2n-1)}{\Gamma(2n+1)}=\int_{0}^{1}(1-x)\,x^{2n-2}\,dx$$ we have: $$\sum_{n=1}^{+\infty}\frac{1}{2n(2n-1)}=\int_{0}^{1}\frac{dx}{1+x}=\log 2.$$