Euler's Proof: Finding the Sum of Reciprocals of Squares Using the Sinc Function
Theorem
$$ \sum_{n =1 }^{\infty} {{1} \over {n^2}} = {{ \pi ^2 } \over { 6 }} $$
Proof
Strategy: This is the solution left by Euler, and it proves the result using nothing other than the Euler representation of the sinc function. The idea is quite fresh and interesting, so once you have seen it, it is more difficult to forget.
Euler representation of the sinc function: $$ {{\sin x} \over {x}} = \prod_{n=1}^{\infty} \left( 1 - {{x^2} \over { \pi^2 n^2}} \right) $$
Expanding the right-hand side of the Euler representation gives the following. $$ \prod_{n=1}^{\infty} \left( 1 - {{x^2} \over { \pi^2 n^2}} \right) = \left( 1 - {{x^2} \over { \pi^2 }} \right) \left( 1 - {{x^2} \over { 4 \pi^2 }} \right) \left( 1 - {{x^2} \over { 9 \pi^2 }} \right) \cdots $$ On the other hand, considering the Maclaurin expansion of the sine function, the sinc function can be expressed as follows. $$ {{\sin x} \over {x}} = {{1} \over {x}} \left( x - {{x^{3}} \over {3!}} + {{x^{5}} \over {5!}} - {{x^{7}} \over {7!}} + \cdots \right) $$ That both sides are equal means that the coefficients of each term are equal, and comparing only the coefficients of $x^{2}$ gives the following. $$ - {{1} \over {3!}} = - {{1} \over {\pi^2 }} - {{1} \over { 4 \pi^2 }} - {{1} \over { 9 \pi^2 }} - \cdots $$ Multiplying both sides by $- \pi^2$ gives $$ {{\pi^2} \over {6}} = {{1} \over { 1 }} + {{1} \over { 4 }} + {{1} \over { 9 }} + \cdots $$ Rearranging gives $$ \sum_{n =1 }^{\infty} {{1} \over {n^2}} = {{ \pi ^2 } \over { 6 }} $$
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