Complex Summation!

Calculus Level 3

∑ n = 1 ∞ 1 n 2 + 9 = π T coth ⁡ ( Q π ) − 1 S \sum_{n=1}^{\infty}{\frac {1}{{n^2}+9}} = \frac{\pi}{T} \coth {\left (Q\pi\right)}-\frac{1}{S}

The above summation holds true for positive integers T T , Q Q and S S . Find T + Q + S T+Q+S .

Inspiration: Aditya Kumar


The answer is 27.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Chew-Seong Cheong
Sep 13, 2018

Relevant wiki: Digamma Function

S = ∑ n = 1 ∞ 1 n 2 + 9 = ∑ n = 1 ∞ 1 ( n − 3 i ) ( n + 3 i ) where i = − 1 is the imaginary unit. = ∑ n = 1 ∞ 1 6 i ( 1 n − 3 i − 1 n + 3 i ) Digamma function ψ 0 ( 1 + z ) = ∑ n = 1 ∞ ( 1 n − 1 z + n ) − γ = 1 6 i ( ψ 0 ( 1 + 3 i ) − ψ 0 ( 1 − 3 i ) ) ψ 0 ( 1 + z ) = ψ 0 ( z ) + 1 z = 1 6 i ( ψ ( 3 i ) + 1 3 i − ψ ( 3 i ) − π cot ⁡ ( 3 π i ) ) and ψ 0 ( 1 − z ) − ψ 0 ( z ) = π cot ⁡ ( π z ) = − 1 18 − π 6 ( e − 3 π + e 3 π e − 3 π − e 3 π ) = π 6 coth ⁡ ( 3 π ) − 1 18 \begin{aligned} S & = \sum_{n=1}^\infty \frac 1{n^2+9} \\ & = \sum_{n=1}^\infty \frac 1{(n-3i)(n+3i)} & \small \color{#3D99F6} \text{where }i = \sqrt{-1} \text{ is the imaginary unit.} \\ & = \sum_{n=1}^\infty \frac 1{6i} \left(\frac 1{n-3i} - \frac 1{n+3i} \right) & \small \color{#3D99F6} \text{Digamma function }\psi_0 (1+z) = \sum_{n=1}^\infty \left(\frac 1n - \frac 1{z+n} \right) - \gamma \\ & = \frac 1{6i} ({\color{#3D99F6}\psi_0(1+3i)}- {\color{#D61F06}\psi_0(1-3i)}) & \small \color{#3D99F6} \psi_0 (1+z) = \psi_0 (z) + \frac 1z \\ & = \frac 1{6i} \left(\color{#3D99F6}\psi(3i) + \frac 1{3i} \color{#D61F06} - \psi(3i) - \pi \cot (3\pi i) \right) & \small \color{#D61F06} \text{ and }\psi_0 (1-z) - \psi_0 (z) = \pi \cot (\pi z) \\ & = - \frac 1{18} - \frac \pi6 \left(\frac {e^{-3\pi}+e^{3\pi}}{e^{-3\pi}-e^{3\pi}}\right) \\ & = \frac \pi 6 \coth (3\pi) - \frac 1{18} \end{aligned}

Therefore, T + S + Q = 6 + 3 + 18 = 27 T+S+Q = 6+3+18= \boxed{27}

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