Leibniz's Formula for Pi/Leibniz's Proof

Theorem

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

That is:

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

Proof

The area $OAT$ is a quarter-circle whose area is $\dfrac \pi 4$ by Area of a Circle.

Now consider the area $C$ of the segment $OPQT$, bounded by the arc $OT$ and the chord $OT$.

Consider the area $OPQ$, bounded by the line segments $OP$ and $OQ$ and the arc $PQ$.

As $P$ and $Q$ approach each other, the arc $PQ$ tends towards the straight line segment $\mathrm d s = PQ$.

We can therefore consider the area $OPQ$ as a triangle.

We extend the line segment $PQ$ and drop a perpendicular $OR$ to $O$.

Using Area of a Triangle in Terms of Side and Altitude, we see that the area $\mathrm d C$ of $\triangle OPQ$ is given by:

$\displaystyle \mathrm d C = \triangle OPQ = \frac {OR \cdot PQ} 2 = \frac {OR \cdot \mathrm d s} 2$

We also note from elementary Euclidean geometry that $\triangle ORS$ is similar to the small triangle on $PQ$.

Thus:

$\displaystyle \frac {\mathrm d s}{\mathrm d x} = \frac {OS} {OR} \iff OR \cdot \mathrm d s = OS \cdot \mathrm d x$

Thus:

$\displaystyle \mathrm d C = \frac {OS \cdot \mathrm d x} 2 = \frac {y \mathrm d x} 2$

where $y = OS$.

We set the horizontal coordinate of $P$ as equal to $x$.

Thus the total area $C$ is equal to the total of all the areas of these small triangles as $x$ increases from $0$ to $1$.

So:

$\displaystyle C = \int \mathrm d C = \frac 1 2 \int_0^1 y \mathrm d x$

Now we use Integration by Parts to swap $x$ and $y$:

$\displaystyle C = \left[{\frac 1 2 x y}\right]_0^1 - \frac 1 2 \int_0^1 x \mathrm d y = \frac 1 2 - \frac 1 2 \int_0^1 x \mathrm d y$

It can be seen that the limits on this new integral have to be $0$ and $1$ from the geometry of the situation.

Now we note that:

• $y = \tan \dfrac \phi 2$
• $x = 1 - \cos \phi = 2 \sin^2 \dfrac \phi 2$ (using the double angle formulas)

Thus:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \tan^2 \frac \phi 2\)	\(=\)	\(\displaystyle \frac {\sin^2 \frac \phi 2}{\cos^2 \frac \phi 2}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of tangent
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sin^2 \frac \phi 2 \sec^2 \frac \phi 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of secant
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sin^2 \frac \phi 2 \left({1 + \tan^2 \frac \phi 2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Sum of Squares of Sine and Cosine: Corollary

and so:

$\displaystyle \frac x 2 = \frac {y^2}{1 + y^2}$

Using Sum of Geometric Progression:

$\displaystyle \frac {y^2}{1 + y^2} = y^2 - y^4 + y^6 - y^8 + \cdots$

This gives us:

$\displaystyle C = \frac 1 2 - \int_0^1 \left({y^2 - y^4 + y^6 - y^8 + \cdots}\right) \mathrm d y$

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle C\)	\(=\)	\(\displaystyle \frac 1 2 - \int_0^1 \left({y^2 - y^4 + y^6 - y^8 + \cdots}\right) \mathrm d y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 2 - \left[{\frac {y^3} 3 - \frac {y^5} 5 + \frac {y^7} 7 - \frac {y^9} 9 + \cdots}\right]_0^1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 2 - \left({\frac 1 3 - \frac 1 5 + \frac 1 7 - \frac 1 9 + \cdots}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 2 - \frac 1 3 + \frac 1 5 - \frac 1 7 + \frac 1 9 - \cdots\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Remember, $C$ is the area of the segment $OPQT$.

Now we add to it the area of $\triangle OTA$, which trivially equals $\dfrac 1 2$, to get the area of the quarter circle which we know as equal to $\dfrac \pi 4$.

Putting it all together, this gives us:

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

$\blacksquare$

Sources

• George F. Simmons: Calculus Gems (1992): Chapter $\text{B}.13$