Leibniz's Formula for Pi/Elementary Proof

Proof
First we note that:


 * $(1): \quad \dfrac 1 {1 + t^2} = 1 - t^2 + t^4 - t^6 + \cdots + t^{4 n} - \dfrac {t^{4 n + 2} } {1 + t^2}$

which is demonstrated here.

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

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


 * $(2): \quad \ds \int_0^x \frac {\d t} {1 + t^2} = x - \frac {x^3} 3 + \frac {x^5} 5 - \frac {x^7} 7 + \cdots + \frac {x^{4 n + 1} } {4 n + 1} - \map {R_n} x$

where:
 * $\ds \map {R_n} x = \int_0^x \frac {t^{4 n + 2} } {1 + t^2} \rd t$

From Square of Real Number is Non-Negative we have that:
 * $t^2 \ge 0$

and so:
 * $1 \le 1 + t^2$

From Relative Sizes of Definite Integrals, we have:
 * $\ds 0 \le \map {R_n} x \le \int_0^x t^{4 n + 2} \rd t$

that is:
 * $0 \le \map {R_n} x \le \dfrac {x^{4n + 3} } {4 n + 3}$

But as $0 \le x \le 1$ it is clear that:
 * $\dfrac {x^{4 n + 3} } {4 n + 3} \le \dfrac 1 {4 n + 3}$

So:
 * $0 \le \map {R_n} x \le \dfrac 1 {4 n + 3}$

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

This leads us directly to:


 * $(3): \quad \ds \int_0^x \frac {\d t} {1 + t^2} = x - \frac {x^3} 3 + \frac {x^5} 5 - \frac {x^7} 7 + \frac {x^9} 9 \cdots$

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

and thence from the Fundamental Theorem of Calculus we have:
 * $\ds \arctan x = \int_0^x \frac {\d t} {1 + t^2}$

From $(3)$ it follows immediately that:
 * $(4): \quad \arctan x = x - \dfrac {x^3} 3 + \dfrac {x^5} 5 - \dfrac {x^7} 7 + \dfrac {x^9} 9 \cdots$

Now all we need to do is plug $x = 1$ into $(4)$.

Comment
Note that we did not just take the Sum of Infinite Geometric Sequence:
 * $\dfrac 1 {1 - \paren {-t^2} } = 1 + \paren {-t^2} + \paren {-t^2}^2 + \paren {-t^2}^3 + \cdots$

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