Did anybody order Digamma?

Calculus Level 2

∑ n = 1 ∞ ( ψ ( 1 2 + n ) − ψ ( 1 2 − n ) ) = ? \sum_{n=1}^\infty \left(\psi \left(\frac 12 + n\right) - \psi \left(\frac 12 - n\right) \right) = \ ?

Notation: ψ ( ⋅ ) \psi (\cdot) denotes the digamma function .


The answer is 0.

Aruna Yumlembam by Aruna Yumlembam , 16, India

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1 solution

Chew-Seong Cheong
May 29, 2020

S = ∑ n = 1 ∞ ( ψ ( 1 2 + n ) − ψ ( 1 2 − n ) ) = ∑ n = 1 ∞ ( ψ ( 1 2 + n ) − ψ ( 1 − ( 1 2 + n ) ) ) Note that ψ ( 1 − z ) − ψ ( z ) = π cot ⁡ ( π z ) = ∑ n = 1 ∞ − π cot ⁡ ( π 2 + n π ) = ∑ n = 1 ∞ π tan ⁡ ( n π ) = 0 \begin{aligned} S & = \sum_{n=1}^\infty \left(\psi \left(\frac 12 + n\right) - \psi \left(\frac 12 - n\right) \right) \\ & = \sum_{n=1}^\infty \left(\psi \left(\frac 12 + n\right) - \psi \left(1 - \left(\frac 12 + n\right)\right) \right) & \small \blue{\text{Note that }\psi (1-z) - \psi (z) = \pi \cot (\pi z)} \\ & = \sum_{n=1}^\infty - \pi \cot \left(\frac \pi 2+n \pi \right) \\ & = \sum_{n=1}^\infty \pi \tan (n \pi) \\ & = \boxed 0 \end{aligned}

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