Leibniz's Formula for Pi/Lemma

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Lemma

$\displaystyle \frac 1 {1 + t^2} = 1 - t^2 + t^4 - t^6 + \cdots + t^{4 n} - \frac {t^{4 n + 2}}{1 + t^2} = \left({\sum_{k \mathop = 0}^{2 n} \left({-1}\right)^k t^{2k}}\right) - \frac {t^{4 n + 2}}{1 + t^2}$

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


Proof

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


From Square of Real Number is Non-Negative, 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 $t$.

$\blacksquare$