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Proof that $\sum\limits_{n=1}^{\infty}\dfrac{1}{n^2}=\dfrac{\pi^2}{6}$
The proof follows Proof 11 of \emph{Evaluating $\zeta(2)$} at
Evaluating $\zeta(2)$
AB3 page 33 shows that if
\[I_n=\int_0^{\pi/2}\sin^nx\,dx\]
then
\[I_{2n}=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\dots\cdot\frac{2n-1}{2n}\cdot\frac{\pi}{2}\]
It follows that
\begin{align}
I_{2n}&=\frac{1\cdot2\cdot3\cdot\dots\cdot(2n-1)\cdot2n}{(2\cdot4\cdot6\cdot\dots\cdot2n)^2}\cdot\frac{\pi}{2}\\\\
&=\frac{1\cdot2\cdot3\cdot\dots\cdot(2n-1)\cdot2n}{(2^n\cdot1\cdot2\cdot3\cdot\dots\cdot n)^2}\cdot\frac{\pi}{2}\\\\
&=\frac{(2n)!}{4^n(n!)^2}\frac{\pi}{2}
\end{align}
(1)
Substituting $x$ by $\pi/2-x$ shows that $I_n=\int_0^{\pi/2}\cos^nx\,dx$. Then use integration of parts to give
\begin{align*}
I_{2n}&=\left[x\cos^{2n}x\right]_0^{\pi/2}+2n\int_0^{\pi/2}x\sin x\cos^{2n-1}x\,dx\\[2ex]
&=n\left[x^2\sin x\cos^{2n-1}x\right]_0^{\pi/2}-n\int_0^{\pi/2}x^2\cos^{2n} x-(2n-1)\sin^2x\cos^{2n-2}x\,dx\\[2ex]
&=n(2n-1)J_{n-1}-2n^2J_n
\end{align*}
where $\displaystyle J_n=\int_0^{\pi/2}x^2\cos^{2n}x\,dx$.
Hence
\[\frac{(2n)!}{4^n(n!)^2}\frac{\pi}{2}=n(2n-1)J_{n-1}-2n^2J_n\]
Multiply both sides by $\dfrac{4^n(n!)^2}{2n^2(2n)!}$ to get
\begin{align*}
\frac{\pi}{4n^2}&=\frac{4^n(n!)^2}{2n^2(2n)!}n(2n-1)J_{n-1}-\frac{4^n(n!)^2}{2n^2(2n)!}2n^2J_n\\\\
&=\frac{4^n}{4(2n)!}\left(\frac{n!}{n}\right)^22n(2n-1)J_{n-1}-\frac{4^n(n!)^2}{(2n)!}J_n\\\\
&=\frac{4^{n-1}((n-1)!)^2}{(2n-2)!}J_{n-1}-\frac{4^n(n!)^2}{(2n)!}J_n
\end{align*}
Now we use the telescoping series trick AA3 page 9 and get
\[\sum_{n=1}^N\frac{\pi}{4n^2}=J_0-\frac{4^N(N!)^2}{(2N)!}J_N\]
We will show that
\[\lim_{N\to\infty}\frac{4^N(N!)^2}{(2N)!}J_N=0\]
from which it will follow that
\[\sum_{n=1}^\infty\frac{\pi}{4n^2}=J_0=\int_0^{\pi/2}x^2\,dx=\frac{\pi^3}{24}\]
so, using the multiple rule for series and multiplying both side by $4/\pi$ we get
\[\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}\]
as we wanted.
\medskip \\
\textbf{To show that}
\[\lim_{N\to\infty}\frac{4^N(N!)^2}{(2N)!}J_N=0\]
we use the \emph{Inequalities for integrals} method of AB3 Section 3.1.
We also use the fact that $x\leq\frac{\pi}{2}\sin x$ for $0 < x < \frac{\pi}{2}$ (proved later) with the Inequality Rule to get
\begin{align*}
J_N&=\int_0^{\pi/2}N^2\cos^{2N}x\,dx\\\\
&\leq\int_0^{\pi/2}\left(\frac{\pi}{2}\sin x\right)^2\cos^{2N}x\,dx\\\\
&=\frac{\pi^2}{4}\int_0^{\pi/2}\sin^2x\cos^{2N}x\,dx\\\\
&=\frac{\pi^2}{4}\int_0^{\pi/2}(1-\cos^2x)\cos^{2N}x\,dx\\\\
&=\frac{\pi^2}{4}\left(\int_0^{\pi/2}\cos^{2N}x\,dx-\int_0^{\pi/2}\cos^{2N+2}x\,dx\right)\\\\
&=\frac{\pi^2}{4}\left(I_{2N}-I_{2N+2}\right)\\\\
&=\frac{\pi^2}{4}\left(I_{2N}-\frac{2N+1}{2N+2}I_{2N}\right)\quad\text{by AB3 Exercise 2.3(b)}\\\\
&=\frac{\pi^2}{4}\frac{1}{2n+2}I_{2N}
\end{align*}
So
\begin{align*}
\frac{4^N(N!)^2}{(2N)!}J_N&\leq\frac{4^N(N!)^2}{(2N)!}\frac{\pi^2}{4}\frac{1}{2n+2}I_{2N}\\\\
&=\frac{\pi^2}{8(N+1)}\frac{4^N(N!)^2}{(2N)!}I_{2N}\\\\
&=\frac{\pi^2}{8(N+1)}\frac{\pi}{2}\quad\text{by (1)}\\\\
&=\frac{\pi^3}{16(N+1)}
\end{align*}
Thus
\[0< J_N\leq\frac{\pi^3}{16}\frac{1}{n+1}\]
By the Squeeze Rule for sequences it follows that
\[\lim_{N\to\infty}\frac{4^N(N!)^2}{(2N)!}J_N=0\]
and we are done.
\medskip \\
\textbf{Finally we must show}
\[x\leq\frac{\pi}{2}\sin x \text{ for }0< x<\frac{\pi}{2}\]
which we can do using Strategy 4.1 of AB2.
Let \[f(x)=\frac{x}{\sin x}\text{ for } 0< x<\frac{\pi}{2}.\]
From now on assume $0< x<\frac{\pi}{2}$
\[f'(x)=\frac{\sin x-x\cos x}{x^2}\geq 0 \text{ as } \cos x\leq\frac{\sin x}{x} \text{ by AB1 page 7}\]
so $f$ is increasing on $(0,\frac{\pi}{2})$ and hence, on this interval,
\[f(x)\leq f\left(\frac{\pi}{2}\right)=\frac{\pi/2}{\sin \pi/2}=\frac{\pi}{2}.\]
Thus \[\frac{x}{\sin x}\leq\frac{\pi}{2}\] so \[x\leq\frac{\pi}{2}\sin x.\]