While proving some results on series I encountered that, one of those result implies that $$\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n)}$$ is convergent and it has sum equal to sum of alternating harmonic series. (And we know that alternating harmonic series converges to $\ln2$.)
However I am not able to find the sum of series $\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n)}$ directly (without that result). Is there is any way to show that sum equals to $\ln2$?
Recall that $$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n}=\ln2$$ Now what we do is part the sum in even and odd parity ie when $n\to 2n $ and $n\to 2n-1$ .$$\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}=\sum_{n\geq 1} \left(\frac{1}{2n-1} -\frac{1}{2n}\right)=\sum_{n\geq 1}\frac{1}{2n(2n-1)}=\ln 2$$
$$\frac{1}{(2n-1)(2n)}=\frac{1}{2n-1}-\frac{1}{2n}$$ So the given sum is nothing but $$S=(1-1/2)+(1/3-1/4)+(1/5-1/6)+(1/7-1/8)+.....+...$$ Now note that $$\ln(1+x)=x-x^2/2+x^3/3-x^4/4+x^5/5+....$$ So $S$ nothing but $x=1$ case of the logarithmic series written above. Therefor $s=\ln 2$
