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Américo Tavares
Américo Tavares
10:32
For instance the accelerated series
\begin{equation*}
\zeta (3)=\sum_{n=1}^{\infty }\frac{1}{n^{3}}=\frac{5}{2}\sum_{1}^{\infty }
\frac{(-1)^{n-1}}{n^{3}\dbinom{2n}{n}}
\end{equation*}
can be obtained by letting $N\rightarrow \infty $ in
\begin{equation*}
\sum_{n=1}^{N}\frac{1}{n^{3}}-2\sum_{n=1}^{N}\frac{\left( -1\right) ^{n-1}}{n^{3}\binom{2n}{n}}=\sum_{k=1}^{N}\frac{(-1)^{k}}{2k^{3}\binom{N+k}{k}\binom{N}{k}}-\sum_{k=1}^{N}\frac{(-1)^{k}}{2k^{3}\binom{2k}{k}}.
\end{equation*}
This last equality can be justified as follows:
(1) write
\begin{equation*}
X_{n,k}=\frac{(-1)^{k-1}}{k^{2}\binom{n+k}{k}\binom{n-1}{k}},\qquad D_{n,k}=\frac{(-1)^{k}}{n^{2}\binom{n+k}{k}\binom{n-1}{k}}\qquad k<n;
\end{equation*}
(2) notice that $X_{n,k}=D_{n,k-1}-D_{n,k}$, and (3) sum over $k$, $1\leq
k\leq n-1$
\begin{equation*}
\sum_{k=1}^{n-1}X_{n,k}=\sum_{k=1}^{n-1}\left( D_{n,k-1}-D_{n,k}\right)
=D_{n,0}-D_{n,n-1}.
\end{equation*}
(4) Finally, sum over $n$, $1\leq n\leq N$, because
\begin{equation*}

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