Proof for Wallis Product

Fun Practice!

Using integration by parts, if $I_{n} = \int_{0}^{π} s i n^{n} θ d θ$ . Prove that $I_{n} = \frac{n - 1}{n} I_{n - 2}$

Let

I_{n} = \int_{0}^{π} \sin^{n} θ d θ

Via integration by parts, solving for $I_{n}$ .
$\int u^{'} v = u v - \int u v^{'}$ where $(u^{'} = \sin θ, v = \sin^{n - 1} θ)$

\begin{aligned} \int \sin^{n} θ d θ & = - \cos θ \sin^{n - 1} θ + \int \cos^{2} θ (n - 1) \sin^{n - 2} θ d θ \\ = - \cos θ \sin^{n - 1} θ + \int (1 - \sin^{2} θ) (n - 1) \sin^{n - 2} θ d θ \\ = - \cos θ \sin^{n - 1} θ + (n - 1) [\int \sin^{n - 2} θ d θ - \int \sin^{n} θ d θ] \\ I_{n} & = - \cos θ \sin^{n - 1} θ + (n - 1) [I_{n - 2} - I_{n}] \\ n I_{n} & = - \cos θ \sin^{n - 1} θ + (n - 1) [I_{n - 2}] \\ I_{n} & = \frac{1}{n} {[- \cos θ \sin^{n - 1} θ + (n - 1) (I_{n - 2})]}_{0}^{π} \\ = \frac{n - 1}{n} I_{n - 2} \end{aligned}

So we proved that

I_{n} = \frac{n - 1}{n} I_{n - 2}

Since $I_{0} = π$ . Thus

I_{2 n} = π \prod_{k = 1}^{n} \frac{2 k - 1}{2 k}

And $I_{1} = 2$

I_{2 n + 1} = 2 \prod_{k = 1}^{n} \frac{2 k}{2 k + 1}

The inequality below holds because $\sin θ \in [- 1, 1]$ which gets smaller as $n$ increases. Thus we are summing up smaller values of $\sin^{n} θ$ as $n$ increases.

I_{2 n + 1} \leq I_{2 n} \leq I_{2 n - 1}

$I_{2 n + 1}$ is given as

I_{2 n + 1} = \frac{2 n}{2 n + 1} I_{2 n - 1}

The limit of $\frac{I_{2 n - 1}}{I_{2 n + 1}}$ as $n$ approaches infinity is

lim_{n \to \infty} \frac{I_{2 n - 1}}{I_{2 n + 1}} = lim_{n \to \infty} \frac{2 n + 1}{2 n} = 1

Using our inequality previously, dividing it throughout by $I_{2 n + 1}$ , and taking the limit of $n$ to $\infty$

1 \leq \frac{I_{2 n}}{I_{2 n + 1}} \leq 1

Thus,

lim_{n \to \infty} \frac{I_{2 n}}{I_{2 n + 1}} = 1

We can thus substitute the previously defined values for $I_{n}$ and $I_{2 n + 1}$

\begin{aligned} \frac{I_{2 n}}{I_{2 n + 1}} & = \frac{π}{2} \prod_{k = 1}^{\infty} \frac{\frac{2 k - 1}{2 k}}{\frac{2 k}{2 k + 1}} \\ 1 & = \frac{π}{2} \prod_{k = 1}^{\infty} \frac{4 k^{2} - 1}{4 k^{2}} \end{aligned}

Thus we have formulated the proof that

π = 2 \prod_{k = 1}^{\infty} (\frac{4 k^{2}}{4 k^{2} - 1})