📂Complex Anaylsis

Computing the Sum of Reciprocals of Squares Using Complex Analysis

Theorem ¹

$$ \sum_{n =1 }^{\infty} {{1} \over {n^2}} = {{ \pi ^2 } \over { 6 }} $$

Euler’s solution is clean and elegant, but the idea is so ingenious that it rarely finds practical use. One of the most enjoyable aspects of studying complex analysis is that shortcuts producing such results appear one after another. Since it also makes for a good example, let us try solving it ourselves.

Proof

If we define $\displaystyle f(z) : = {{1} \over {z^2}}$, then $\displaystyle \lim_{z \to \infty} z f(z) = 0$.

Formula for the sum of a series over all integers: For a rational function $f$, suppose that $\lim_{n \to \infty} z f(z) = 0, n \in \mathbb{Z}$ with $f(n) \ne 0$. When $f$ has finitely many singularities $z_{1}, \cdots , z_{m}$, $$ \sum_{n=-\infty}^{\infty} f(n) = - \sum_{n = 1}^{m} \text{Res}_{z_{n}} (\pi f(z) \cot \pi z) $$

It cannot be applied directly; we need to handle the exception for $n \ne 0$.

If we define $F(z): = \pi f(z) \cot \pi z$, then for $n \ne 0$ we have $\text{Res}_{n} F(z) = f(n)$.

Laurent expansion of the cotangent: $$ \cot z = {{1} \over {z}} - {{z} \over {3}} - {{z^{3}} \over {45}} - {{2 z^{5}} \over {945}} - \cdots \\ \csc z = {{1} \over {z}} + {{z} \over {6}} + {{7 z^{3}} \over {360}} + {{31 z^{5}} \over {15120}} + \cdots $$

On the other hand, in the neighborhood of $n=0$, $$ F(z) = {{\pi} \over {z^2}} \left( {{1} \over { \pi z}} - {{ \pi z} \over {3}} - {{ \pi^{3} z^{3}} \over {45}} - \cdots \right) $$ so $$ \text{Res}_{0} F(z) = - {{\pi^2} \over {3}} $$ Since $F$ has no singularities other than at $n \in \mathbb{Z}$, $$ \sum_{n=-\infty}^{\infty} f(n) = \sum_{n=-\infty}^{-1} f(n) - {{\pi^2} \over {3}} + \sum_{n=1}^{\infty} f(n) = 0 $$ Since $f$ is an even function, $\displaystyle \sum_{n=-\infty}^{-1} f(n) = \sum_{n=1}^{\infty} f(n)$, and rearranging gives $$ \sum_{n =1 }^{\infty} {{1} \over {n^2}} = {{ \pi ^2 } \over { 6 }} $$

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See Also

• Euler’s solution