Series Expansion for Pi over Root 2/Proof 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
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 | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac \pi {\sqrt 2}\) |
$\blacksquare$