Leibniz's Formula for Pi/Lemma

Lemma

$\ds \frac 1 {1 + t^2}$

$=$

$\ds 1 - t^2 + t^4 - t^6 + \cdots + t^{4 n} - \frac {t^{4 n + 2} } {1 + t^2}$

$\ds $

$=$

$\ds \paren {\sum_{k \mathop = 0}^{2 n} \paren {-1}^k t^{2 k} } - \frac {t^{4 n + 2} } {1 + t^2}$

This holds for all real $t \in \R$.

$\ds \frac {1 - \paren {-t^2}^{2 n + 1} } {1 - \paren {-t^2} }$

$=$

$\ds \sum_{k \mathop = 0}^{2 n} \paren {-t^2}^k$

$\ds \leadsto \ \ $

$\ds \frac {1 + \paren {t^2}^{2 n + 1} } {1 + t^2}$

$=$

$\ds 1 - t^2 + t^4 - t^6 + \cdots + t^{4 n}$

$\ds \leadsto \ \ $

$\ds \frac 1 {1 + t^2} + \frac {t^{4 n + 2} } {1 + t^2}$

$=$

$\ds 1 - t^2 + t^4 - t^6 + \cdots + t^{4 n}$

$\ds \leadsto \ \ $

$\ds \frac 1 {1 + t^2}$

$=$

$\ds 1 - t^2 + t^4 - t^6 + \cdots + t^{4 n} - \frac {t^{4 n + 2} } {1 + t^2}$

$t^2 \ge 0$

for all real $t$.

So $- t^2 \le 0$ and so $- t^2 \ne 1$.

So the conditions of Sum of Geometric Sequence are satisfied, and so the above argument holds for all real $t$.

$\blacksquare$