A Beauuuuu Tiful series

Calculus Level 3

If 1 π 2 + 1 + 1 4 π 2 + 1 + 1 9 π 2 + 1 + 1 16 π 2 + 1 + ⋯ = 1 e p − q \dfrac{1}{\pi^2+1}+\dfrac{1}{4\pi^2+1}+ \dfrac{1}{9\pi^2+1}+\dfrac{1}{16\pi^2+1}+\cdots = \dfrac{1}{e^p-q} where p p and q q are positive integers, find the value of ( p + q ) 2 (p+q)^2 .


The answer is 9.

Naren Bhandari by Naren Bhandari , 21, India

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1 solution

Chew-Seong Cheong
Nov 20, 2018

Relevant wiki: Digamma Function

S = ∑ n = 1 ∞ 1 ( n π ) 2 + 1 = ∑ n = 1 ∞ 1 ( 1 + n π i ) ( 1 − n π i ) = 1 2 ∑ n = 1 ∞ ( 1 1 + n π i + 1 1 − n π i ) = i 2 π ∑ n = 1 ∞ ( 1 n + i π − 1 n − i π ) By ψ ( 1 + z ) = − γ + ∑ n = 1 ∞ ( 1 n − 1 n + z ) = i 2 π ( ψ ( 1 − i π ) − ψ ( 1 + i π ) ) where ψ ( ⋅ ) denotes the digamma function. = i 2 π ( π cot ⁡ ( i ) + i π ) Since ψ ( 1 − z ) = ψ ( z ) + π cot ⁡ ( π z ) = 1 2 ( e + e − 1 e − e − 1 − 1 ) and ψ ( 1 + z ) = ψ ( z ) + 1 z = 1 e 2 − 1 \begin{aligned} S & = \sum_{n=1}^\infty \frac 1{(n\pi)^2+1} \\ & = \sum_{n=1}^\infty \frac 1{(1+n\pi i)(1-n\pi i)} \\ & = \frac 12 \sum_{n=1}^\infty \left(\frac 1{1+n\pi i} + \frac 1{1-n\pi i}\right) \\ & = \frac i{2\pi} \sum_{n=1}^\infty \left(\frac 1{n+ \frac i\pi} - \frac 1{n - \frac i\pi}\right) & \small \color{#3D99F6} \text{By }\psi (1+z) = - \gamma + \sum_{n=1}^\infty \left(\frac 1n - \frac 1{n+z}\right) \\ & = \frac i{2\pi} \left(\psi \left(1-\frac i\pi\right) - \psi \left(1+\frac i\pi\right) \right) & \small \color{#3D99F6} \text{where }\psi (\cdot) \text{ denotes the digamma function.} \\ & = \frac i{2\pi} \left(\pi \cot(i) + i\pi \right) & \small \color{#3D99F6} \text{Since }\psi (1-z) = \psi(z) + \pi \cot(\pi z) \\ & = \frac 12 \left(\frac {e + e^{-1}}{e - e^{-1}} -1 \right) & \small \color{#3D99F6} \text{and }\psi (1+z) = \psi(z) + \frac 1z \\ & = \frac 1{e^2-1} \end{aligned}

Therefore, ( p + q ) 2 = ( 2 + 1 ) 2 = 9 (p+q)^2 = (2+1)^2 = \boxed 9 .

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Sir, In the second line you are missing i i .

Naren Bhandari - 2 years, 6 months ago

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Thanks. I got it changed.

Chew-Seong Cheong - 2 years, 6 months ago

@Naren Bhandari Hello......!! Another interesting question.....but you can probably look at this question

Aaghaz Mahajan - 2 years, 6 months ago

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