A calculus problem by dimitris kouroupos

$\begin{aligned} S & = \sum_{n=1}^\infty \frac 1{(2n+1)(2n+3)} \\ & = \frac 12 \sum_{n=1}^\infty \left( \frac 1{2n+1}- \frac 1{2n+3} \right) \\ & = \frac 12 \left( \frac 13 \cancel{-\frac 15 + \frac 15} \cancel{-\frac 17 + \frac 17} \cancel{-\frac 19 + \frac 19} + \ ... \right) \\ & = \boxed{\dfrac 16} \end{aligned}$

If a, b real numbers so as to apply :

• $\frac{1}{ (2n+1)*(2n+3)}$ = $\frac{a}{2n+1}$ + $\frac{b}{2n+3}$ => $1=a(2n+3)+b(2n+1)=>$ $1=n(2a+2b)+(3a+b)=>$ $\large \begin{cases} \ 2a \ + \ 2b\ =0 \\ 3a+b=1 \\ \end{cases}$ $=>$ $\large \begin{cases} \ a \ =\dfrac 12 \ \\ b=-\dfrac 12 \\ \end{cases}$

• therefore

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2n+1)*(2n+3)}$ = $\displaystyle \sum_{n=1}^{\infty}\frac{a+b}{(2n+1)*(2(n+1)+1)}$ = $\frac{1}{2}$ $(\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2n+1)}$ - $\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2(n+1)+1)} )$ = $\frac{1}{2}$ ( $\frac{1}{3}$ - $\large \lim _{n\to{\infty} }\left(\frac{1}{2n+1}\right).)$ = $\frac{1}{6}$