📂Complex Anaylsis

Summation of Inverse Squares Using Complex Analysis

Theorem ¹

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

Euler’s approach is neat and elegant, but it’s so ingenious that there aren’t many practical applications. The joy of studying complex analysis is that such shortcuts to results are readily available. It’s a good example, so try solving it yourself.

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 series over all integers: For the rational function $f$, let’s say $\lim_{n \to \infty} z f(z) = 0, n \in \mathbb{Z}$ where $f(n) \ne 0$. When $f$ has a finite singularity $z_{1}, \cdots , z_{m}$, $$ \sum_{n=-\infty}^{\infty} f(n) = - \sum_{n = 1}^{m} \text{Res}_{z_{n}} (\pi f(z) \cot \pi z) $$

It can’t be directly applied completely, and exception handling for $n \ne 0$ is necessary.

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

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

Meanwhile, in the vicinity of $n=0$, $$ F(z) = {{\pi} \over {z^2}} \left( {{1} \over { \pi z}} - {{ \pi z} \over {3}} - {{ \pi^{3} z^{3}} \over {45}} - \cdots \right) $$ thus $$ \text{Res}_{0} F(z) = - {{\pi^2} \over {3}} $$ $F$ doesn’t have singularities other than $n \in \mathbb{Z}$, so $$ \sum_{n=-\infty}^{\infty} f(n) = \sum_{n=-\infty}^{-1} f(n) - {{\pi^2} \over {3}} + \sum_{n=1}^{\infty} f(n) = 0 $$ Since $f$ is an even function, $\displaystyle \sum_{n=-\infty}^{-1} f(n) = \sum_{n=1}^{\infty} f(n)$ and if we organize it, we get $$ \sum_{n =1 }^{\infty} {{1} \over {n^2}} = {{ \pi ^2 } \over { 6 }} $$

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See Also

• Euler’s approach