proving $\sum_{n=0}^{\infty }\frac{1}{(2n+1)(2n+2)}=\sum_{n=0}^{\infty }\frac{4}{(4n+1)(4n+2)(4n+3)}$ I want to prove $$\sum_{n=0}^{\infty }\frac{1}{(2n+1)(2n+2)}=\sum_{n=0}^{\infty }\frac{4}{(4n+1)(4n+2)(4n+3)}$$ 
 A: \begin{eqnarray}
\sum_{n=0}^{\infty}\frac{1}{(2n+1)(2n+2)}
&=&\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}
=\sum_{n=0}^{\infty}(\frac{1}{4n+1}-\frac{1}{4n+2}+\frac{1}{4n+3}-\frac{1}{4n+4})\\
&=&\sum_{n=0}^{\infty}(\frac{1}{4n+1}-\frac{2}{4n+2}+\frac{1}{4n+3})
+\sum_{n=0}^{\infty}(\frac{1}{4n+2}-\frac{1}{4n+4})\\
&=&\sum_{n=0}^{\infty}\frac{2}{(4n+1)(4n+2)(4n+3)}
+\frac{1}{2}\sum_{n=0}^{\infty}\frac{1}{(2n+1)(2n+2)}
\end{eqnarray}
A: $$\frac{1}{(2n+1)(2n+2)}=\frac{1}{2n+1}-\frac{1}{2n+2},$$
hence:
$$\sum_{n=0}^{+\infty}\frac{1}{(2n+1)(2n+2)}=\sum_{n\geq 1}\frac{(-1)^{n-1}}{n}=\int_{0}^{1}\sum_{n\geq 0}(-x)^{n}\,dx=\int_{0}^{1}\frac{dx}{1+x}=\log 2,$$
while:
$$\frac{4}{(4n+1)(4n+2)(4n+3)}=\frac{2}{4n+1}-\frac{2}{2n+1}+\frac{2}{4n+3}$$
so:
$$\begin{eqnarray*}\sum_{n\geq 0}\frac{4}{(4n+1)(4n+2)(4n+3)}&=&2\int_{0}^{1}\sum_{n\geq 0}(x^{4n}-2x^{4n+1}+x^{4n+2})\,dx\\&=&2\int_{0}^{1}\frac{(1-x)^2}{1-x^4}\,dx\\&=&2\int_{0}^{1}\frac{1-x}{(1+x)(1+x^2)}\,dx\end{eqnarray*}$$
and by setting $x=\frac{1-y}{1+y}$ we get that the last integral equals:
$$ \int_{0}^{1}\frac{2y}{y^2+1}\,dy=\log 2.$$
