Trigamma summation

Calculus Level 5

∑ n = 1 ∞ ψ 1 ( n ) 2 n − 1 = π A B + ( ln ⁡ C ) D \large \sum _{ n=1 }^{ \infty }{ \dfrac { { \psi }_{ 1 }( n ) }{ { 2 }^{ n-1 } } } =\dfrac { { \pi }^{ A } }{ B } +{ ( \ln { C } ) }^{ D }

The above equation holds true for positive integers A , B , C A,B, C and D D . Find A + B + C + D A+B+C+D .

Notation : ψ 1 ( ⋅ ) \psi_1 (\cdot) denotes the 1 st 1^\text{st} derivative of the Digamma function .


The answer is 12.

Aditya Kumar by Aditya Kumar , 22, India

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1 solution

Otto Bretscher
May 5, 2016

ψ 1 ( n ) = ζ ( 2 ) − H n ( 2 ) + 1 n 2 \psi_1(n)=\zeta(2)-H_n^{(2)}+\frac{1}{n^2} so ∑ n = 1 ∞ ψ 1 ( n ) x n = ζ ( 2 ) x 1 − x − L i 2 ( x ) 1 − x + L i 2 ( x ) \sum_{n=1}^{\infty}\psi_1(n)x^n=\zeta(2)\frac{x}{1-x}-\frac{Li_2(x)}{1-x}+Li_2(x) . Evaluating at x = 1 2 x=\frac{1}{2} and multiplying with 2 gives π 2 6 + ( ln ⁡ 2 ) 2 \frac{\pi^2}{6}+(\ln2)^2 so that the answer is 12 \boxed{12}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

As intended! Can you add the generalization of ψ A ( x ) { \psi }_{ A }\left( x \right) ? It will help new users.

Aditya Kumar - 5 years, 1 month ago

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I'm a little busy at work. We have an analogous formula ψ n ( x ) = ( − 1 ) n + 1 n ! ( ζ ( n + 1 ) − H x − 1 ( n + 1 ) ) \psi_n(x)=(-1)^{n+1}n!\left(\zeta(n+1)-H_{x-1}^{(n+1)}\right) , I believe... the interested user will be able to work out the series as an exercise ;)

Otto Bretscher - 5 years, 1 month ago

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