## Série 1/(1⋅3) + 1/(3⋅5) + 1/(5⋅7) + ⋯ = 1/2

Desta questão de Jack Thompson, no MSE. Demonstrar que

$\dfrac{1}{1 \cdot 3} + \dfrac{1}{3 \cdot 5} + \dfrac{1}{5 \cdot 7} + \cdots = \dfrac{1}{2}.$

Tradução da minha resolução (utilizando a pista de David Mitra, em um comentário).

Escrever a série na forma telescópica e calcular a sua soma:

\begin{aligned}S&=\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\cdots \\ &=\sum_{n=1}^{\infty }\frac{1}{\left( 2n-1\right) \left( 2n+1\right) }\\&=\sum_{n=1}^{\infty }\left( \frac{1}{2\left( 2n-1\right) }-\frac{1}{2\left( 2n+1\right) }\right)\quad\text{Decomposi\c{c}\~{a}o em frac\c{c}\~{o}es parciais}\\&=\frac{1}{2}\sum_{n=1}^{\infty }\left( \frac{1}{2n-1}-\frac{1}{2n+1}\right) \qquad \text{S\'{e}rie telesc\'{o}pica}\\&=\frac{1}{2}\sum_{n=1}^{\infty }\left( a_{n}-a_{n+1}\right), \qquad a_{n}=\frac{1}{2n-1},a_{n+1}=\frac{1}{2\left( n+1\right) -1}=\frac{1}{2n+1}\\&=\frac{1}{2}\left( a_{1}-\lim_{n\rightarrow \infty }a_{n}\right) \qquad\text{veja em baixo}\\&=\frac{1}{2}\left( \frac{1}{2\cdot 1-1}-\lim_{n\rightarrow \infty }\frac{1}{2n-1}\right)\\&=\frac{1}{2}\left( 1-0\right)\\&=\frac{1}{2}.\end{aligned}

A soma da série telescópica $\displaystyle\sum_{n=1}^{\infty }\left( a_{n}-a_{n+1}\right)$ é o limite da soma telescópica $\displaystyle\sum_{n=1}^{N}\left( a_{n}-a_{n+1}\right)$. Dado que

\begin{aligned}\displaystyle\sum_{n=1}^{N}\left( a_{n}-a_{n+1}\right)&=\left( a_{1}-a_{2}\right)+\left( a_{2}-a_{3}\right) +\ldots +\left( a_{N-1}-a_{N}\right) +\left( a_{N}-a_{N+1}\right)\\&=a_{1}-a_{2}+a_{2}-a_{3}+\ldots +a_{N-1}-a_{N}+a_{N}-a_{N+1}\\&=a_{1}-a_{N+1},\end{aligned}

tem-se

\begin{aligned}\displaystyle\sum_{n=1}^{\infty }\left( a_{n}-a_{n+1}\right)&=\lim_{N\rightarrow\infty}\displaystyle\sum_{n=1}^{N}\left( a_{n}-a_{n+1}\right)\\&=\lim_{N\rightarrow \infty }\left( a_{1}-a_{N+1}\right)\\&=a_{1}-\lim_{N\rightarrow \infty}a_{N+1}\\&=a_{1}-\lim_{N\rightarrow \infty }a_{N}\\&=a_{1}-\lim_{n\rightarrow \infty }a_{n}.\end{aligned}