Leibniz's Formula for Pi

Theorem

 * $$\frac \pi 4 = 1 - \frac 1 3 + \frac 1 5 - \frac 1 7 + \frac 1 9 - \cdots $$

That is:
 * $$\pi = 4 \sum_{k \ge 0} \left({-1}\right)^k \frac 1 {2 k + 1}$$

Proof
First we note that:


 * $$(1) \qquad \frac 1 {1+t^2} = 1 - t^2 + t^4 - t^6 + \cdots + t^{4n} - \frac {t^{4n + 2}}{1+t^2}$$

which is demonstrated here.

Now consider the real number $$x \in \R: 0 \le x \le 1$$.

We can integrate expression $$(1)$$ WRT $$t$$ from $$0$$ to $$x$$:


 * $$(2) \qquad \int_{0}^{x}\frac {\mathrm{d}{t}}{1+t^2} = x - \frac {x^3} 3 + \frac {x^5} 5 - \frac {x^7} 7 + \cdots + \frac {x^{4n + 1}}{4n + 1} = R_n \left({x}\right)$$

where:
 * $$R_n \left({x}\right) = \int_{0}^{x} \frac {t^{4n + 2}}{1+t^2}\mathrm{d}{t}$$

From Even Powers are Positive we have that $$t^2 \ge 0$$ and so $$1 \le 1 + t^2$$.

So:
 * $$0 \le R_n \left({x}\right) \le \int_{0}^{x} t^{4n + 2} \mathrm{d}{t}$$

that is:
 * $$0 \le R_n \left({x}\right) \le \frac {x^{4n + 3}}{4n + 3}$$

But as $$0 \le x \le 1$$ it is clear that $$\frac {x^{4n + 3}}{4n + 3} \le \frac 1 {4n + 3}$$.

So:
 * $$0 \le R_n \left({x}\right) \le \frac 1 {4n + 3}$$

From Basic Null Sequences and the Squeeze Theorem, $$\frac 1 {4n + 3} \to 0$$ as $$n \to \infty$$.

This leads us directly to:


 * $$(2) \qquad \int_{0}^{x}\frac {\mathrm{d}{t}}{1+t^2} = x - \frac {x^3} 3 + \frac {x^5} 5 - \frac {x^7} 7 + \frac {x^9} 9 \cdots$$

But from Derivative of Arctangent Function, we also have that:
 * $$\frac{\mathrm{d}{}}{\mathrm{d}{x}} \arctan t = \frac 1 {1+t^2}$$

and thence from the Fundamental Theorem of Calculus we have:
 * $$\arctan x = \int_{0}^{x}\frac {\mathrm{d}{t}}{1+t^2}$$

From $$(2)$$ it follows immediately that:
 * $$(3) \qquad \arctan x = x - \frac {x^3} 3 + \frac {x^5} 5 - \frac {x^7} 7 + \frac {x^9} 9 \cdots$$

Now all we need to do is plug $$x = 1$$ into $$(3)$$.

Comment
Note that we did not just take the Sum of Infinite Geometric Progression:
 * $$\frac 1 {1 - \left({-t^2}\right)} = 1 + \left({-t^2}\right) + \left({-t^2}\right)^2 + \left({-t^2}\right)^3 + \cdots$$

and integrate it term by term, as we have not at this stage proved that this is permissible.

Alternative proof
We obtain $$(2)$$ above directly from Taylor Expansion of Arctangent Function.