Leibniz's Formula for Pi/Lemma

Lemma

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

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

Proof

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac {1 - \left({-t^2}\right)^{2n+1} } {1 - \left({-t^2}\right)}\)	\(=\)	\(\displaystyle \sum_{k=0}^{2 n} \left({-t^2}\right)^k\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Sum of Geometric Progression
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \frac {1 + \left({t^2}\right)^{2n+1} } {1 + t^2}\)	\(=\)	\(\displaystyle 1 - t^2 + t^4 - t^6 + \cdots + t^{4n}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \frac 1 {1 + t^2} + \frac {t^{4n + 2} } {1 + t^2}\)	\(=\)	\(\displaystyle 1 - t^2 + t^4 - t^6 + \cdots + t^{4n}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \frac 1 {1 + t^2}\)	\(=\)	\(\displaystyle 1 - t^2 + t^4 - t^6 + \cdots + t^{4n} - \frac {t^{4n + 2} } {1 + t^2}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

From Even Powers are Positive, we have that $t^2 \ge 0$ for all real $t$.

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

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

$\blacksquare$