Series Expansion for Pi over Root 2

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Theorem

$\ds \frac \pi {\sqrt 2} = \sum_{r \mathop = 1}^\infty \paren {-1}^{r - 1} \frac {r - \frac 1 2} {r^2 - r + \frac 3 {16} }$


Proof 1

Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as:

$\map f x = \begin{cases} \sin \dfrac x 2 & : 0 \le x < \pi \\ -\sin \dfrac x 2 & : \pi < x \le 2 \pi \end{cases}$


From Half-Range Fourier Sine Series: $\sin \dfrac x 2$ over $\closedint 0 \pi$, $-\sin \dfrac x 2$ over $\closedint 0 {2 \pi}$, we have:

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


Setting $x = \dfrac {\pi} 2$, we have:

\(\ds \sin \dfrac \pi 4\) \(=\) \(\ds \frac 8 \pi \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {n \sin \dfrac {n \pi} 2 } {4 n^2 - 1}\)
\(\ds \leadsto \ \ \) \(\ds \frac {\sqrt 2} 2\) \(=\) \(\ds \frac 8 \pi \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {n \sin \dfrac {n \pi} 2 } {4 n^2 - 1}\) Sine of $\dfrac \pi 4$
\(\ds \leadsto \ \ \) \(\ds \frac \pi {\sqrt 2}\) \(=\) \(\ds 8 \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {n \sin \dfrac {n \pi} 2 } {4 n^2 - 1}\) rearranging


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

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

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


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

Hence we have:

\(\ds \frac \pi {\sqrt 2}\) \(=\) \(\ds 8 \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {n \sin \dfrac {n \pi} 2 } {4 n^2 - 1}\) from above
\(\ds \) \(=\) \(\ds 8 \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {n \sin \dfrac {n \pi} 2 } {\paren {2 n + 1} \paren {2 n - 1} }\) Difference of Two Squares
\(\ds \) \(=\) \(\ds 8 \sum_{r \mathop = 1}^\infty \paren {-1}^{\paren {2 r - 1} - 1} \frac {\paren {2 r - 1} \sin \dfrac {\paren {2 r - 1} \pi} 2 } {\paren {2 \paren {2 r - 1} + 1} \paren {2 \paren {2 r - 1} - 1} }\) from above: terms in even $n$ vanish
\(\ds \) \(=\) \(\ds 8 \sum_{r \mathop = 1}^\infty \paren {-1}^{2 r} \frac {\paren {2 r - 1} \map \sin {\paren {r - 1} + \frac 1 2} \pi} {\paren {4 r - 1} \paren {4 r - 3} }\) simplification
\(\ds \) \(=\) \(\ds 8 \sum_{r \mathop = 1}^\infty \frac {\paren {2 r - 1} \map \sin {\paren {r - 1} + \frac 1 2} \pi} {\paren {4 r - 1} \paren {4 r - 3} }\) $\paren {-1}^{2 r} = 1$
\(\ds \) \(=\) \(\ds 8 \sum_{r \mathop = 1}^\infty \frac {\paren {2 r - 1} \paren {-1}^{r - 1} } {\paren {4 r - 1} \paren {4 r - 3} }\) Sine of Half-Integer Multiple of Pi
\(\ds \) \(=\) \(\ds 8 \sum_{r \mathop = 1}^\infty \paren {-1}^{r - 1} \frac {2 \paren {r - \frac 1 2} } {16 r^2 - 16 r + 3}\) multiplying out
\(\ds \) \(=\) \(\ds \sum_{r \mathop = 1}^\infty \paren {-1}^{r - 1} \frac {r - \frac 1 2} {r^2 - r + \frac 3 {16} }\) simplifying

$\blacksquare$


Proof 2

We have:

\(\ds \frac {r - \frac 1 2} {r^2 - r + \frac 3 {16} }\) \(=\) \(\ds \frac {16 r - 8} {16 r^2 - 16 r + 3}\)
\(\ds \) \(=\) \(\ds 2 \paren {\frac {8 r - 4} {\paren {4 r - 1} \paren {4 r - 3} } }\) factorising, extracting a factor of $2$
\(\ds \) \(=\) \(\ds 2 \paren {\frac {\paren {4 r - 3} + \paren {4 r - 1} } {\paren {4 r - 1} \paren {4 r - 3} } }\)
\(\ds \) \(=\) \(\ds 2 \paren {\frac 1 {4 r - 1} + \frac 1 {4 r - 3} }\)

So, we have:

\(\ds \sum_{r \mathop = 1}^\infty \paren {-1}^r \frac {r - \frac 1 2} {r^2 - r + \frac 3 {16} }\) \(=\) \(\ds -2 \sum_{r \mathop = 1}^\infty \paren {-1}^r \paren {\frac 1 {4 r - 1} + \frac 1 {4 r - 3} }\)
\(\ds \) \(=\) \(\ds -2 \sum_{r = 1}^\infty \paren {-1}^r \paren {\int_0^1 x^{4 r - 2} \rd x + \int_0^1 x^{4 r - 4} \rd x}\) Primitive of Power, Fundamental Theorem of Calculus
\(\ds \) \(=\) \(\ds -2 \sum_{r \mathop = 1}^\infty \paren {-1}^r \int_0^1 \paren {x^{4 r - 2} + x^{4 r - 4} } \rd x\) Linear Combination of Definite Integrals
\(\ds \) \(=\) \(\ds -2 \int_0^1 \paren {\sum_{r \mathop = 1}^\infty \paren {-1}^r \paren {x^{4 r - 2} + x^{4 r - 4} } } \rd x\) Fubini's Theorem
\(\ds \) \(=\) \(\ds -2 \int_0^1 \paren {\frac 1 {x^2} \sum_{r \mathop = 1}^\infty \paren {-x^4}^r - \sum_{r \mathop = 0}^\infty \paren {-x^4}^r} \rd x\) shifting the index in the second sum, and extracting a factor of $x^{-4}$ from the first
\(\ds \) \(=\) \(\ds -2 \int_0^1 \paren {\frac 1 {x^2} \paren {\sum_{r \mathop = 0}^\infty \paren {-x^4}^r - 1} - \sum_{r \mathop = 0}^\infty \paren {-x^4}^r} \rd x\)
\(\ds \) \(=\) \(\ds -2 \int_0^1 \paren {\paren {\frac 1 {x^2} - 1} \sum_{r \mathop = 0}^\infty \paren {-x^4}^r - \frac 1 {x^2} } \rd x\) grouping terms
\(\ds \) \(=\) \(\ds -2 \int_0^1 \paren {\paren {\frac {1 - x^2} {x^2} } \paren {\frac 1 {1 + x^4} } - \frac 1 {x^2} } \rd x\) Sum of Infinite Geometric Sequence
\(\ds \) \(=\) \(\ds -2 \int_0^1 \paren {\frac {1 - x^2 - \paren {1 + x^4} } {x^2 \paren {1 + x^4} } } \rd x\) writing as a single fraction
\(\ds \) \(=\) \(\ds 2 \int_0^1 \paren {\frac {x^2 \paren {1 + x^2} } {x^2 \paren {1 + x^4} } } \rd x\) extracting a factor of $x^2$
\(\ds \) \(=\) \(\ds 2 \int_0^1 \frac {1 + x^2} {1 + x^4} \rd x\) cancelling the factor of $x^2$
\(\ds \) \(=\) \(\ds \sqrt 2 \intlimits {\map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } } 0 1\) Primitive of $\dfrac {1 + x^2} {1 + x^4}$
\(\ds \) \(=\) \(\ds \sqrt 2 \paren {\arctan 0 - \lim_{x \mathop \to 0^+} \map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } }\)
\(\ds \) \(=\) \(\ds -\sqrt 2 \lim_{u \mathop \to -\infty} \arctan u\) setting $\dfrac 1 {\sqrt 2} \paren {x - \frac 1 x} = u$, Arctangent of Zero is Zero
\(\ds \) \(=\) \(\ds -\sqrt 2 \paren {-\frac \pi 2}\) Limit to Infinity of Arctangent Function
\(\ds \) \(=\) \(\ds \frac \pi {\sqrt 2}\)

$\blacksquare$

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