1 Introduction
This gives step by step the solution Euler came up to show that
\begin{align*} \sum _{n=1}^{\infty }\frac {1}{n^{2}} & =1+\frac {1}{2^{2}}+\frac {1}{3^{2}}+\cdots \\ & =\frac {\pi ^{2}}{6}\end{align*}
2 Solution steps
Starting with the Taylor series of \(\sin x\)
\[ \sin x=x-\frac {x^{3}}{3!}+\frac {x^{5}}{5!}-\cdots \]
We want only non-zero roots. But
\(\sin x\) has root at
\(x=0\). Hence dividing
both sides by
\(x\) gives
\begin{equation} \frac {\sin x}{x}=1-\frac {x^{2}}{3!}+\frac {x^{4}}{5!}-\cdots \tag {A}\end{equation}
The left side is the
\(\operatorname {sinc}\left ( x\right ) \) which has no zero root. All its roots are at integer multiple
of
\(\pi \). Using Fundamental theory of algebra which says for finite polynomial of degree
\(n\) it can be
written as product of its
\(n\) roots, then the above becomes
\begin{equation} \frac {\sin x}{x}=\cdots \left ( 1-\frac {x}{-3\pi }\right ) \left ( 1-\frac {x}{-2\pi }\right ) \left ( 1-\frac {x}{-\pi }\right ) \left ( 1-\frac {x}{\pi }\right ) \left ( 1-\frac {x}{2\pi }\right ) \left ( 1-\frac {x}{3\pi }\right ) \cdots \tag {1}\end{equation}
Euler assumed this applies to
infinite polynomial as it does for finite polynomial. It took 100 years for Karl Weierstrass
to shows this is true with his Weierstrass factorization theorem. (1) can be rewritten
as
\begin{align} \frac {\sin x}{x} & =\overbrace {\left ( 1+\frac {x}{\pi }\right ) \left ( 1-\frac {x}{\pi }\right ) }\overbrace {\left ( 1+\frac {x}{2\pi }\right ) \left ( 1-\frac {x}{2\pi }\right ) }\overbrace {\left ( 1+\frac {x}{3\pi }\right ) \left ( 1-\frac {x}{3\pi }\right ) }\cdots \tag {2}\\ & =\left ( 1-\frac {1}{\pi ^{2}}x^{2}\right ) \left ( 1-\frac {1}{4\pi ^{2}}x^{2}\right ) \left ( 1-\frac {1}{9\pi ^{2}}x^{2}\right ) \cdots \tag {3}\end{align}
Where each term in (3) is the product of each two terms under brackets in (2). Hence (3) becomes
(adding two more terms to the end)
\[ \frac {\sin x}{x}=\left ( 1-\frac {1}{\pi ^{2}}x^{2}\right ) \left ( 1-\frac {1}{4\pi ^{2}}x^{2}\right ) \left ( 1-\frac {1}{9\pi ^{2}}x^{2}\right ) \left ( 1-\frac {1}{16\pi ^{2}}x^{2}\right ) \left ( 1-\frac {1}{25\pi ^{2}}x^{2}\right ) \cdots \]
Multiplying throughout gives
\begin{align*} \frac {\sin x}{x} & =\overbrace {\left ( 1-\frac {1}{\pi ^{2}}x^{2}\right ) \left ( 1-\frac {1}{4\pi ^{2}}x^{2}\right ) }\left ( 1-\frac {1}{9\pi ^{2}}x^{2}\right ) \left ( 1-\frac {1}{16\pi ^{2}}x^{2}\right ) \left ( 1-\frac {1}{25\pi ^{2}}x^{2}\right ) \cdots \\ & =\overbrace {\left ( 1-\frac {1}{4\pi ^{2}}x^{2}-\frac {1}{\pi ^{2}}x^{2}+\frac {x^{4}}{4\pi ^{4}}\right ) \left ( 1-\frac {1}{9\pi ^{2}}x^{2}\right ) }\left ( 1-\frac {1}{16\pi ^{2}}x^{2}\right ) \left ( 1-\frac {1}{25\pi ^{2}}x^{2}\right ) \cdots \\ & =\overbrace {\left ( 1-\frac {1}{9\pi ^{2}}x^{2}+\allowbreak \frac {1}{36\pi ^{4}}x^{4}-\frac {1}{4\pi ^{2}}x^{2}+\allowbreak \frac {1}{9\pi ^{4}}x^{4}-\frac {1}{\pi ^{2}}x^{2}+\allowbreak \frac {1}{4\pi ^{4}}x^{4}-\frac {1}{36\pi ^{6}}x^{6}\right ) }\left ( 1-\frac {1}{25\pi ^{2}}x^{2}\right ) \cdots \end{align*}
And so on. Collecting only the terms with \(x^{2}\) gives
\begin{equation} \frac {\sin x}{x}=\left [ x^{2}\left ( -\frac {1}{\pi ^{2}}-\frac {1}{4\pi ^{2}}-\frac {1}{9\pi ^{2}}-\frac {1}{16\pi ^{2}}-\cdots \right ) \right ] +x^{4}\left ( \cdots \right ) +x^{6}\left ( \cdots \right ) +\cdots \tag {4}\end{equation}
But from (A) we had
\[ \frac {\sin x}{x}=1-\frac {x^{2}}{3!}+\frac {x^{4}}{5!}-\cdots \]
Hence comparing
coefficients of
\(x^{2}\) shows that
\[ -\frac {1}{\pi ^{2}}-\frac {1}{4\pi ^{2}}-\frac {1}{9\pi ^{2}}-\frac {1}{16\pi ^{2}}-\cdots =-\frac {1}{3!}\]
Hence
\begin{align*} \frac {1}{\pi ^{2}}+\frac {1}{4\pi ^{2}}+\frac {1}{9\pi ^{2}}+\frac {1}{16\pi ^{2}}+\cdots & =\frac {1}{6}\\ 1+\frac {1}{4}+\frac {1}{9}+\frac {1}{16}+\cdots & =\frac {\pi ^{2}}{6}\end{align*}
Or
\[ \sum _{n=1}^{\infty }\frac {1}{n^{2}}=\frac {\pi ^{2}}{6}\]
The above derivation made Euler very famous at the time as this was not known result. The
same process can be used to find
\(\sum _{n=1}^{\infty }\frac {1}{n^{4}}\) and for all other even powers. Currently there is no
known result for odd powers. But it was proved that for
\(\sum _{n=1}^{\infty }\frac {1}{n^{3}}\) the sum must be irrational
number.
3 References
- https://en.wikipedia.org/wiki/Basel_problem wikipedia article on Basel
problem.
- https://www.math.cmu.edu/~bwsulliv/basel-problem.pdf Basel problem
different proofs by Sullivan.
- https://web.williams.edu/Mathematics/sjmiller/public_html/hudson/Emmell,%20Amber_Euler%20&%20The%20Basel%20Problem.pdf
Emmell article on Basel problem.