Exercise 8.3.13

Question

\begin{enumerate}[label=(\alph*)] 
\item Show
$$\int_0^{\pi/2} \theta d\theta = \sum^\infty_{n=0} \frac{c_n}{2n + 1} b_{2n+1},$$
being careful to justify each step in the argument. The term $b_{2n+1}$ refers back to our earlier work on Wallis's product.
\item Deduce
$$\frac{\pi^2}{8} = \sum^\infty_{n=0} \frac{1}{(2n+1)^2},$$
and use this to finish the proof that $\pi^2 / 6 = \sum^\infty_{n=1} 1/n^2$.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item From the substitution of $x = \sin(\theta)$  in (5), integrate both sides:
$$\int_0^{\pi/2} \theta d\theta = \int_0^{\pi/2}  \left( \lim_{N \to \infty}\sum^N_{n=0} \frac{c_n}{2n + 1}\sin^{2n+1} \theta \right) d\theta$$
Since the series converges uniformly, by the Integrable Limit Theorem (Theorem 7.4.4) we can move the integral inside of the limit, and since integrals preserve addition, we can move the integral inside of the summation as well:
$$\int_0^{\pi/2} \theta d\theta = \lim_{N \to \infty}  \sum^N_{n=0}\left(  \frac{c_n}{2n + 1} \int_0^{\pi/2}\sin^{2n+1} \theta d\theta \right) = \sum^\infty_{n=0} \frac{c_n}{2n+1}b_{2n+1} $$
\item The left side evaluates to $\pi^2/8$. Recall
$$b_{2n+1} = \prod^n_{i=1} \frac{2i}{2i+1} \text{ and } c_n = \prod^n_{i=1} \frac{2i-1}{2i}$$
so
$$\frac{c_n}{2n+1} b_{2n+1} = \frac{1}{2n+1} \prod^n_{i=1} \frac{2i-1}{2i+1}$$
The product telescopes and is equal to $1/(2n+1)$, so
$$\frac{\pi^2}{8} = \sum^\infty_{n=0} \frac{1}{(2n+1)^2}$$
Note that $\sum^\infty_{n=1} 1/n^2$ and $\sum^\infty_{n=0}1 / (2n+1)^2$ both converge absolutely, so we're free to reorder terms as necessary. Let $a = \sum^\infty_{n=1} 1/n^2$, and note that

We have
$$\sum^\infty_{n=1} \frac{1}{n^2} = \sum^\infty_{n=1} \frac{1}{(2n-1)^2} + \sum^\infty_{n=1} \frac{1}{(2n)^2} = \sum^\infty_{n=0} \frac{1}{(2n+1)^2} + \frac{1}{4}\sum^\infty_{n=1} \frac{1}{n^2}$$
A little algebra on this result shows $\pi^2 / 6 = \sum^\infty_{n=1} 1/n^2$.
 \end{enumerate}

Exercise 8.3.13

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Exercise 8.3.13

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