Series Expansion for Pi over 8 Root 2

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Theorem

$\displaystyle \frac \pi {8 \sqrt 2} = \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {2 n - 1} {\paren {4 n - 1} \paren {4 n - 3} }$


Proof

Let $\map f x$ be the real function defined on $\openint {-\pi} \pi$ as:

$\map f x = \begin{cases} \cos x & : -\pi < x < 0 \\ -\cos x & : 0 < x < \pi \end{cases}$


From Fourier Series: $\cos x$ over $\openint {-\pi} 0$, $-\cos x$ over $\openint 0 \pi$, we have:

$\displaystyle \map f x \sim -\frac 8 \pi \sum_{r \mathop = 1}^\infty \frac {r \sin 2 r x} {4 r^2 - 1}$


Setting $x = \dfrac \pi 4$, we have:

\(\displaystyle -\cos \dfrac \pi 4\) \(=\) \(\displaystyle -\frac 8 \pi \sum_{r \mathop = 1}^\infty \frac {r \sin 2 r \dfrac \pi 4} {4 r^2 - 1}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \cos \dfrac \pi 4\) \(=\) \(\displaystyle \frac 8 \pi \sum_{r \mathop = 1}^\infty \frac {r \sin \dfrac {r \pi} 2} {4 r^2 - 1}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\sqrt 2} 2\) \(=\) \(\displaystyle \frac 8 \pi \sum_{r \mathop = 1}^\infty \frac {r \sin \dfrac {r \pi} 2} {4 r^2 - 1}\) Cosine of $\dfrac \pi 4$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\pi} {8 \sqrt 2}\) \(=\) \(\displaystyle \sum_{r \mathop = 1}^\infty \frac {r \sin \dfrac {r \pi} 2} {4 r^2 - 1}\) rearranging


When $r$ is even, $\dfrac {r \pi} 2$ is an integer multiple of $\pi$.

Hence, in this case, from Sine of Multiple of Pi:

$\sin \dfrac {r \pi} 2 = 0$


When $r$ is odd it can be expressed as $r = 2 n - 1$ for $n \ge 1$.

Hence we have:

\(\displaystyle \frac {\pi} {8 \sqrt 2}\) \(=\) \(\displaystyle \sum_{r \mathop = 1}^\infty \frac {r \sin \dfrac {r \pi} 2} {4 r^2 - 1}\) from above
\(\displaystyle \) \(=\) \(\displaystyle \sum_{r \mathop = 1}^\infty \frac {r \sin \dfrac {r \pi} 2} {\paren {2 r + 1} \paren {2 r - 1} }\) Difference of Two Squares
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac {\paren {2 n - 1} \sin \dfrac {\paren {2 n - 1} \pi} 2} {\paren {2 \paren {2 n - 1} + 1} \paren {2 \paren {2 n - 1} - 1} }\) from above: terms in even $r$ vanish
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac {\paren {2 n - 1} \sin \paren {\paren {n - 1} + \frac 1 2} \pi} {\paren {4 n - 1} \paren {4 n - 3} }\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac {\paren {2 n - 1} \paren {-1}^{n - 1} } {\paren {4 n - 1} \paren {4 n - 3} }\) Sine of Half-Integer Multiple of Pi
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {2 n - 1} {\paren {4 n - 1} \paren {4 n - 3} }\) rearranging

$\blacksquare$


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