## Fórmulas de Apéry

Prove as seguintes fórmulas

1. $\displaystyle\sum_{k=1}^{K}\dfrac{a_{1}a_{2}...a_{k-1}}{(x+a_{1})(x+a_{2})...(x+a_{k})}=\dfrac{1}{x}-\dfrac{a_{1}a_{2}...a_{K}}{x(x+a_{1})(x+a_{2})...(x+a_{K})}$$\bigskip$
2. Caso particular para $x=n^{2}$ e $a_{k}=-k^{2}$
$\displaystyle\sum_{k=1}^{n-1}\dfrac{\left( -1\right) ^{k-1}\left( k-1\right) !^{2}}{\left( n^{2}-1^{2}\right) ...\left( n^{2}-k^{2}\right) }=\dfrac{1}{n^{2}}-\frac{\left( -1\right) ^{n-1}\left( n-1\right) !^{2}}{n^{2}\left( n^{2}-1^{2}\right) ...\left[ n^{2}-\left( n-1\right) ^{2}\right] }$$\bigskip$
3. $\displaystyle\sum_{n=1}^{N}\displaystyle\sum_{k=1}^{n-1}(-1)^{k}\left( \varepsilon _{n,k}-\varepsilon_{n-1,k}\right) =2\left( \displaystyle\sum_{n=1}^{N}\dfrac{1}{n^{3}}-\dfrac{2\left( -1\right) ^{n-1}}{n^{3}\dbinom{2n}{n}}\right)$
em que
$\displaystyle\varepsilon _{n,k}=\dfrac{1}{k^{3}\dbinom{n}{k}\dbinom{n+k}{k}}$$\bigskip$
4. $\displaystyle\sum_{n=1}^{N}\displaystyle\sum_{k=1}^{n-1}\left( -1\right) ^{k}\left( \varepsilon_{n,k}-\varepsilon _{n-1,k}\right)=\displaystyle\sum_{k=1}^{N}\displaystyle\sum_{n=k+1}^{N}\left( -1\right) ^{k}\left( \varepsilon _{n,k}-\varepsilon _{n-1,k}\right)=$ $=\displaystyle\sum_{k=1}^{N}\left( -1\right) ^{k}\left( \varepsilon _{N,k}-\varepsilon_{k,k}\right)$$\bigskip$
5. $\displaystyle\sum_{k=1}^{N}\left( -1\right) ^{k}\left( \varepsilon _{N,k}-\varepsilon_{k,k}\right)=\displaystyle\sum_{k=1}^{N}\dfrac{\left( -1\right) ^{k}}{k^{3}\dbinom{N}{k}\dbinom{N+k}{k}}-\displaystyle\sum_{k=1}^{N}\dfrac{\left( -1\right) ^{k}}{k^{3}\dbinom{2k}{k}}$$\bigskip$
6. $\displaystyle\sum_{n=1}^{N}\dfrac{1}{n^{3}}+\displaystyle\sum_{k=1}^{N}\dfrac{\left( -1\right) ^{k-1}}{2k^{3}\dbinom{N}{k}\dbinom{N+k}{k}}=$ $\dfrac{5}{2}\displaystyle\sum_{k=1}^{N}\frac{\left( -1\right) ^{k-1}}{k^{3}\dbinom{2k}{k}}$

Resolução: veja a secção 3 deste artigo de Alf van der Poorten

A proof that Euler Missed … (também disponível  aqui), ou veja estas minhas notas.