An unusual sum

Calculus Level 5

∑ r = 1 2014 ψ ( r 2015 ) sin ⁡ ( 2 π r 2015 ) = − A B π \large{\displaystyle \sum _{ r=1 }^{ 2014 }{ \psi \left( \frac { r }{ 2015 } \right) \sin { \left( \frac { 2\pi r }{ 2015 } \right) } } =-\frac{A}{B}\pi}

Given that the summation above is equal to − A B π -\dfrac AB \pi , where A A and B B are coprime positive integers, and ψ ( x ) \psi (x) denote the digamma function. Find the value of A + B A+B ,


The answer is 2015.

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Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

It is known that ( see more ):

∑ r = 1 m − 1 ψ ( r m ) sin ⁡ ( 2 k π r m ) = π 2 ( 2 k − m ) ⇒ ∑ r = 1 2014 ψ ( r 2015 ) sin ⁡ ( 2 π r 2015 ) = π 2 ( 2 − 2015 ) = − 2013 2 π \begin{aligned} \sum_{r=1}^{m-1} \psi \left(\frac{r}{m}\right) \sin \left(\frac{2k\pi r}{m}\right) & = \frac{\pi}{2}(2k-m) \\ \Rightarrow \sum_{r=1}^{2014} \psi \left(\frac{r}{2015}\right) \sin \left(\frac{2\pi r}{2015}\right) & = \frac{\pi}{2}(2-2015) = -\frac{2013}{2} \pi \end{aligned}

⇒ A + B = 2013 + 2 = 2015 \Rightarrow A+B = 2013+2 = \boxed{2015}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Wow! I didn't know these identities. Upvoted +1. Do you know how to prove it? @Tanishq Varshney and @Chew-Seong Cheong ?

Pi Han Goh - 5 years, 6 months ago

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