Stoneham's series

[Thomas Klemm]

In both cases we can use the infinite product representation of the sinc function:

\(
\begin{align}
\frac {\sin \pi z}{\pi z}=\prod _{n=1}^{\infty }\left(1-{\frac {z^{2}}{n^{2}}}\right)
\end{align}
\)

For \(z=\frac{1}{2}\) we have:

\(
\begin{align}
\frac {\sin \pi z}{\pi z}
&=\frac {\sin \pi \frac{1}{2}}{\pi \frac{1}{2}} \\
\\
&=\frac{2}{\pi} \\
\\
&=\prod _{n=1}^{\infty }\left(1-{\frac {1}{(2n)^2}}\right) \\
\\
&=\prod _{n=1}^{\infty }\frac{(2n-1)(2n+1)}{(2n)(2n)} \\
\\
&=\frac{1 \cdot 3}{2 \cdot 2} \cdot \frac{3 \cdot 5}{4 \cdot 4} \cdot \frac{5 \cdot 7}{6 \cdot 6} \cdots \\
\end{align}
\)

And thus:

\(
\begin{align}
\frac{\pi}{2}&=\frac{2 \cdot 2}{1 \cdot 3} \cdot \frac{4 \cdot 4}{3 \cdot 5} \cdot \frac{6 \cdot 6}{5 \cdot 7} \cdots \\
\\
\pi
&=4 \cdot \frac{2 \cdot 4}{3 \cdot 3} \cdot \frac{4 \cdot 6}{5 \cdot 5} \cdot \frac{6 \cdot 8}{7 \cdot 7} \cdots \\
\\
&=4 \cdot \left(1 - \frac{1}{3^2}\right) \cdot \left(1 - \frac{1}{5^2}\right) \cdot \left(1 - \frac{1}{7^2}\right) \cdots \\
\end{align}
\)

Similarly for \(z=\frac{1}{4}\) we have:

\(
\begin{align}
\frac {\sin \pi z}{\pi z}
&=\frac {\sin \pi \frac{1}{4}}{\pi \frac{1}{4}} \\
\\
&=\frac{2 \sqrt{2}}{\pi} \\
\\
&=\prod _{n=1}^{\infty }\left(1-{\frac {1}{(4n)^2}}\right) \\
\\
&=\prod _{n=1}^{\infty }\frac{(4n-1)(4n+1)}{(4n)(4n)} \\
\\
&=\frac{3 \cdot 5}{4 \cdot 4} \cdot \frac{7 \cdot 9}{8 \cdot 8} \cdot \frac{11 \cdot 13}{12 \cdot 12} \cdots \\
\end{align}
\)

If we multiply both products we get:

\(
\begin{align}
\frac{\pi}{2} \cdot \frac{2 \sqrt{2}}{\pi}
&= \sqrt{2} \\
\\
&=\frac{2 \cdot 2}{1 \cdot 3} \cdot \frac{6 \cdot 6}{5 \cdot 7} \cdot \frac{10 \cdot 10}{9 \cdot 11} \cdots \\
\\
\sqrt{2} = \frac{2}{\sqrt{2}}
&= 2 \cdot \frac{1 \cdot 3}{2 \cdot 2} \cdot \frac{5 \cdot 7}{6 \cdot 6} \cdot \frac{9 \cdot 11}{10 \cdot 10} \cdots \\
\\
&=2 \cdot \left(1 - \frac{1}{4 \cdot 1^2}\right) \cdot \left(1 - \frac{1}{4 \cdot 3^2}\right) \cdot \left(1 - \frac{1}{4 \cdot 5^2}\right) \cdots \\
\end{align}
\)

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This might remind you of the Wallis product for \(\pi\):
Proof using Euler's infinite product for the sine function