How can we prove this identity:∑n=1∞arctan⁡(−1)nϕ2n+1=∑n=1∞(−1)n5,ϕ4n−2,(2n−1),F2n−1where Fn are the Fibonacci numbers, and ϕ=1+52 is the golden ratio?

By using the power series expansion of arctan⁡(x) at x=0 (note that |(−1)nϕ2n+1|&lt;1), we have that∑n=1∞arctan⁡(−1)nϕ2n+1=∑n=1∞∑k=0∞(−1)n+kϕ−(2k+1)(2n+1)2k+1=∑k=0∞(−1)kϕ−(2k+1)2k+1∑n=1∞(−ϕ−(4k+2))n=∑k=0∞(−1)kϕ−(2k+1)2k+1⋅−ϕ−(4k+2)1+ϕ−(4k+2)=∑k=0∞(−1)k+1ϕ−(6k+3)2k+1⋅ϕ2k+1ϕ(2k+1)−(−ϕ−1)2k+1=∑k=0∞(−1)k+1ϕ−(4k+2)2k+1⋅15F2k+1=15∑n=1∞(−1)nϕ−(4n−2)(2n−1)F2n−1

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markush35q

Answered question

2022-03-24

How can we prove this identity:

\sum_{n = 1}^{\infty} \arctan \frac{{(- 1)}^{n}}{ϕ^{2 n + 1}} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{\sqrt{5}, ϕ^{4 n - 2}, (2 n - 1), F_{2 n - 1}}

where

F_{n}

are the Fibonacci numbers, and

ϕ = \frac{1 + \sqrt{5}}{2}

is the golden ratio?

Answer & Explanation

Nunnaxf18

Beginner2022-03-25Added 18 answers

By using the power series expansion of

\arctan (x)

x = 0

(note that

| \frac{{(- 1)}^{n}}{ϕ^{2 n + 1}} | < 1

), we have that

\sum_{n = 1}^{\infty} \arctan \frac{{(- 1)}^{n}}{ϕ^{2 n + 1}} = \sum_{n = 1}^{\infty} \sum_{k = 0}^{\infty} {(- 1)}^{n + k} \frac{ϕ^{- (2 k + 1) (2 n + 1)}}{2 k + 1}

= \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} ϕ^{- (2 k + 1)}}{2 k + 1} \sum_{n = 1}^{\infty} {(- ϕ^{- (4 k + 2)})}^{n}

= \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} ϕ^{- (2 k + 1)}}{2 k + 1} \cdot \frac{- ϕ^{- (4 k + 2)}}{1 + ϕ^{- (4 k + 2)}}

= \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k + 1} ϕ^{- (6 k + 3)}}{2 k + 1} \cdot \frac{ϕ^{2 k + 1}}{ϕ^{(2 k + 1)} - {(- ϕ^{- 1})}^{2 k + 1}}

= \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k + 1} ϕ^{- (4 k + 2)}}{2 k + 1} \cdot \frac{1}{\sqrt{5} F_{2 k + 1}}

= \frac{1}{\sqrt{5}} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} ϕ^{- (4 n - 2)}}{(2 n - 1) F_{2 n - 1}}

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