Resolved: "For a∈R, show that ∑k≥01k2+k−α=π4α+1tan⁡(12π4α+1)"

Carly Shannon

Answered question

2022-01-23

For

a \in R

, show that

\sum_{k \geq 0} \frac{1}{k^{2} + k - α} = \frac{π}{\sqrt{4 α + 1}} \tan (\frac{1}{2} π \sqrt{4 α + 1})

Aiden Cooper

Beginner2022-01-24Added 14 answers

First we note that

S = \sum_{k \geq 0} \frac{1}{k^{2} + k - α} = \sum_{k \geq 0} \frac{1}{{(k + \frac{1}{2})}^{2} - \frac{4 a + 1}{4}} = 4 \sum_{k \geq 0} \frac{1}{{(2 k + 1)}^{2} - a^{2}}

where

a = \sqrt{4 α + 1}

, and so

S = 4 \sum_{m o d d} \frac{1}{m^{2} - a^{2}}

Now we use the well-known contangent identity

π \cot (a π) = \frac{1}{a} + \sum_{m = 1}^{\infty} \frac{2 a}{a^{2} - m^{2}}

(1)
Replacing a by a/2 and dividing by 2 gives

\frac{1}{2} π \cot (\frac{a π}{2}) = \frac{1}{a} + \sum_{m = 1}^{\infty} \frac{2 a}{a^{2} - {(2 m)}^{2}}

(2)
Subtracting (2) from (1) gives

π \cot (a π) - \frac{1}{2} π \cot (\frac{a π}{2}) = \sum_{m = 1, o d d m}^{\infty} \frac{2 a}{a^{2} - m^{2}}

but

\cot (θ) - \frac{1}{2} \cot (\frac{θ}{2}) = - \frac{1}{2} \tan (\frac{θ}{2})

and so

\frac{π}{4 a} \tan (\frac{a π}{2}) = \sum_{m o d d} \frac{1}{m^{2} - a^{2}}

from which the result follows setting

a = \sqrt{4 α + 1}

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