# Pi as Sum of Odd Reciprocals Alternating in Sign in Pairs

## Theorem

$\dfrac {\pi \sqrt 2} 4 = 1 + \dfrac 1 3 - \dfrac 1 5 - \dfrac 1 7 + \dfrac 1 9 + \dfrac 1 {11} - \dfrac 1 {13} - \dfrac 1 {15} \cdots$

## Proof

Let $f: \R \to \R$ be the real function defined as:

$\forall x \in \R: \map f x = \dfrac {x^1} 1 + \dfrac {x^3} 3 - \dfrac {x^5} 5 - \dfrac {x^7} 7 + \dfrac {x^9} 9 + \dfrac {x^{11} } {11} - \dfrac {x^{13} } {13} - \dfrac {x^{15} } {15} \cdots$

We first confirm that the series will converge at $x = 1$.

By grouping the series two terms at a time, we are guaranteed convergence by the Alternating Series Test:

$\map f 1 = \paren {1 + \dfrac 1 3 } - \paren { \dfrac 1 5 + \dfrac 1 7 } + \paren { \dfrac 1 9 + \dfrac 1 {11} } - \cdots + \paren {-1}^{\floor {\frac {k - 1} 2} } \paren {\dfrac 1 {2k - 1 } + \dfrac 1 {2k + 1 } }$

We now move on to determining the sum of $\map f 1$.

We have:

 $\ds \map f x$ $=$ $\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {\frac {x^{4 n + 1} } {4 n + 1} + \frac {x^{4 n + 3} } {4 n + 3} }$ $\ds \leadsto \ \$ $\ds \map {f'} x$ $=$ $\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {x^{4 n} + x^{4 n + 2} }$ Power Rule for Derivatives $\ds$ $=$ $\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {1 + x^2} \paren {x^{4 n} }$ Product of Powers $\ds$ $=$ $\ds \paren {1 + x^2} \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {x^{4 n} }$ $\ds$ $=$ $\ds \paren {1 + x^2} \sum_{n \mathop = 0}^\infty \paren {-x^4 }^n$ Power of Power and Power of Product $\ds$ $=$ $\ds \dfrac {1 + x^2} {1 - \paren{-x^4 } }$ Sum of Infinite Geometric Sequence $\ds$ $=$ $\ds \dfrac {1 + x^2} {1 + x^4}$ $\ds \leadsto \ \$ $\ds \map f 1$ $=$ $\ds \int_0^1 \dfrac {1 + x^2} {1 + x^4} \rd x$ Fundamental Theorem of Calculus/Second Part $\ds$ $=$ $\ds \intlimits {\dfrac 1 {\sqrt 2} \map \arctan {\dfrac 1 {\sqrt 2} \paren {x - \dfrac 1 x} } } 0 1$ Primitive of One plus x Squared over One plus Fourth Power of x $\ds$ $=$ $\ds \dfrac 1 {\sqrt 2} \map \arctan {\dfrac 1 {\sqrt 2} \paren {1 - \dfrac 1 1} } - \dfrac 1 {\sqrt 2} \map \arctan {\dfrac 1 {\sqrt 2} \paren {0 - \dfrac 1 0} }$ $\ds$ $=$ $\ds \dfrac 1 {\sqrt 2} \paren {\map \arctan 0 - \map \arctan {-\infty} }$ $\ds$ $=$ $\ds \dfrac 1 {\sqrt 2} \paren {0 - \paren {-\dfrac \pi 2} }$ Definition of Real Arctangent $\ds$ $=$ $\ds \dfrac {\pi \sqrt 2} 4$

$\blacksquare$