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\font\ninerm=amr9
\centerline{\bf An Elementary Sum\footnote*{\ninerm A nice little proof of
a beautiful, well known theorem. This theorem was proved in 1736
by Leonhard Euler (1707--1783).}}
\vskip 15 pt
\nopagenumbers
\centerline{\sl We show that $\sum_{n=1}^\infty{1\over{n^2}}={{\pi^2}\over{6}}$
,}
\centerline{\sl using only elementary trigonometry and algebra!}
\vskip 15pt

\noindent
For the moment fix $n>0$ and for $1\leq
k \leq n$ set $\theta_k = {{k\pi}\over{(2n + 1)}}$.
The first step is to use De~Moivre's
formula to construct a polynomial whose roots are $\cot^2(\theta_k), k = 1, 
\dots , n$.  Recall that
$$\eqalign{\sin[(2n+1)\theta]&=\Im(e^{(2n+1)i\theta})\cr
  &=\Im\{[\cos(\theta) + i \sin(\theta)]^{2n+1}\}\cr
  &=\sum_{k=0}^n(-1)^k{{2n+1}\choose{2k+1}}\sin^{2k+1} (\theta)\cos^{2(n-k)}
(\theta)\cr
  &=\Bigl[\sum_{k=0}^n (-1)^k{{2n+1}\choose{2k+1}}\cot^{2(n-k)}(\theta)
\Bigr]\Bigl[\sin^{2n+1}(\theta)\Bigr]\cr}$$
Since $\sin({{k\pi}\over{2n+1}})\neq 0$ for $k=1,\dots,n$, the roots of
$p(x)=\sum_{k=0}^n{{2n+1}\choose{2k+1}} (-1)^kx^{n-k}$ are exactly $\cot^2(
\theta_k)$.

For any polynomial $p(x)=a_nx^n+a_{n-1} x^{n-1} + \dots + a_0$, the sum of the
roots is equal to ${a_{n-1}}/{a_n}$.  Therefore,
$$\sum_{k=1}^n \cot^2(\theta_k)={{{2n+1}\choose{3}}\over{{2n+1}\choose{1}}}=
{{(2n+1)2n(2n-1)}\over{3\cdot2\cdot(2n+1)}}={{n(2n-1)}\over{3}}.$$
$$\sum_{k=1}^n\csc^2(\theta_k)=\sum_{k=1}^n 1-\cot^2(\theta_k)=
{{2(n+1)n}\over3}.
$$
Also on $[0,1]$, we know that $\tan(x) \geq x \geq \sin(x)$. Thus,
$$\tan(\theta_k)\geq \theta_k \geq \sin(\theta_k)$$
$$\cot^2(\theta_k)\leq {1\over{\theta_k^2}}\leq\csc^2(\theta_k)$$
$${{n(2n-1)}\over{3}}\leq\sum_{k=1}^n{1\over{\theta_k^2}}
\leq{{2n(n+1)}\over3}$$
$${{\pi^2 n(2n-1)}\over{3(2n+1)^2}}\leq\sum_{k=1}^n{1\over{k^2}}\leq
{{\pi^2 2n(n+1)}\over{3(2n+1)^2}}$$
An application of the sandwich theorem completes the proof:
$$\lim_{n\to\infty}\sum_{k=1}^n {1\over{k^2}}={{\pi^2}\over6}.$$
\bye
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