Deduza

 \zeta(3)=\displaystyle\sum_{k=1}^{\infty}\dfrac{1}{k^3}=\frac{5}{2}\sum_{k=1}^{\infty}\frac{\left(-1\right) ^{k-1}}{k^{3}\dbinom{2k}{k}},

 

a partir de

\displaystyle\sum_{n=1}^{N}\frac{1}{n^{3}}+\sum_{k=1}^{N}\frac{\left( -1\right) ^{k-1}}{2k^{3}\dbinom{N}{k}\dbinom{N+k}{k}}=\frac{5}{2}\sum_{k=1}^{N}\frac{\left(-1\right) ^{k-1}}{k^{3}\dbinom{2k}{k}}

SUGESTÃO: faça tender N para infinito.