Answered: "I want to show that∑k=1n−111−cos⁡(2kπn)=n2−16"

embeucadaoo8

Answered question

2022-02-23

I want to show that

\sum_{k = 1}^{n - 1} \frac{1}{1 - \cos (\frac{2 k π}{n})} = \frac{n^{2} - 1}{6}

Elsie Tillman

Beginner2022-02-24Added 5 answers

z = e^{i t}, z \neq 1

, then

\cos t = \frac{1}{2} (z + z^{- 1})

hence

\frac{1}{1 - \cos t} = - 2 \frac{z}{{(1 - z)}^{2}}

. The sum

s_{n}

to be computed is

s_{n} = \sum_{z}^{\cdot} \frac{- 2 z}{{(1 - z)}^{2}}

where

\sum_{z}^{\cdot}

means that the sum runs over every n-th root z of 1 different from 1. For every

| x | < 1

, consider

t_{n} (x) = \sum_{z} \frac{z x}{{(1 - z x)}^{2}}

where

\sum_{z}

means that the sum runs over every n-th root z of 1. Expanding each ratio

z \frac{x}{{(1 - z x)}^{2}}

as a power series in (zx), one gets

t_{n} (x) = \sum_{k \geq 1} k u_{k}^{n} x^{k}, u_{k}^{n} = \sum_{z} z^{k}

Now,

u_{k}^{n} = 0

for every k except when k is a multiple of n, in which case

u_{k}^{n} = n

. Thus

t_{n} (x) = n^{2} \sum_{k \geq 1} k x^{n k} = \frac{n^{2} x^{n}}{{(1 - x^{n})}^{2}}

This allows to compute

s_{n}

since

s_{n} = 2 \cdot lim_{x \to 1} (\frac{x}{{(1 - x)}^{2}} - t_{n} (x))

Expanding both terms in the limit in powers of 1-x when

x \to 1

yields finally

s_{n} = 2 \cdot \frac{n^{2} - 1}{12} = \frac{n^{2} - 1}{6}

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