Proof of the Basel Problem

1. Power Series for $\sin^{-1}(x)$

Step 1: Differentiate

\[\frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1 - x^2}}.\]

Step 2: Expand using the binomial theorem

\[\frac{1}{\sqrt{1 - x^2}} = \sum_{n=0}^{\infty} \frac{(2n)!}{4^n (n!)^2} x^{2n}.\]

Step 3: Integrate term-by-term

\[\sin^{-1}(x) = \sum_{n=0}^{\infty} \frac{(2n)!}{4^n (n!)^2 (2n+1)} x^{2n+1}.\]

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2. Expressing the Integral as a Series

Let

\[I = \int_0^1 \frac{\sin^{-1}(x)}{\sqrt{1 - x^2}} \, dx.\]

Substitute the series for $\sin^{-1}(x)$:

\[I = \int_0^1 \frac{1}{\sqrt{1 - x^2}} \left( \sum_{n=0}^{\infty} \frac{(2n)!}{4^n (n!)^2 (2n+1)} x^{2n+1} \right) dx.\]

Interchanging the sum and the integral:

\[I = \sum_{n=0}^{\infty} \frac{(2n)!}{4^n (n!)^2 (2n+1)} \int_0^1 \frac{x^{2n+1}}{\sqrt{1 - x^2}} \, dx.\]

Trigonometric substitution:

Let $x = \sin \theta$, then $dx = \cos \theta \, d\theta$, and the integral becomes:

\[\int_0^1 \frac{x^{2n+1}}{\sqrt{1 - x^2}} dx = \int_0^{\pi/2} \sin^{2n+1} \theta \, d\theta.\]

Using Wallis’ integral formula:

\[\int_0^{\pi/2} \sin^{2n+1} \theta \, d\theta = \frac{(2n)!!}{(2n+1)!!} = \frac{4^n (n!)^2}{(2n+1)!}.\]

Substitute back:

\[I = \sum_{n=0}^{\infty} \frac{(2n)!}{(2n+1)(2n+1)!} = \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}.\]

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3. Proof of the Basel Problem

We want to prove:

\[\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}.\]

From the previous result:

\[\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = I = \frac{\pi^2}{8}.\]

Split full sum:

\[\sum_{n=1}^{\infty} \frac{1}{n^2} = \left( \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} \right) + \left( \sum_{k=1}^{\infty} \frac{1}{(2k)^2} \right).\]

Note:

\[\sum_{k=1}^{\infty} \frac{1}{(2k)^2} = \sum_{k=1}^{\infty} \frac{1}{4k^2} = \frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{k^2}.\]

Let $S = \sum_{n=1}^{\infty} \frac{1}{n^2}$, then:

\[S = \frac{\pi^2}{8} + \frac{1}{4} S \Rightarrow \frac{3}{4} S = \frac{\pi^2}{8} \Rightarrow S = \frac{4}{3} \cdot \frac{\pi^2}{8} = \frac{\pi^2}{6}.\]

$\blacksquare$