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 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: