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Prove $$2\sum_{k=1}^n{(-1)^{k-1} \cos\frac{k\pi}{2n+1}} = 1$$

I got this question when I was going through some basic trigonometric identities as follows

$2(\cos\frac{\pi}{5}-\cos\frac{2\pi}{5}) =1\tag1$

very much straightforward to recognise

$2(\cos\frac{\pi}{7}-\cos\frac{2\pi}{7} + \cos\frac{3\pi}{7})=1\tag2$

Following steps proves the identity

$8\cos\frac{\pi}{7}\cdot\cos\frac{2\pi}{7}\cdot\cos\frac{4\pi}{7} = -1$

$4\cos\frac{\pi}{7}[2\cos\frac{4\pi}{7}\cdot\cos\frac{2\pi}{7}] = -1$

$4\cos\frac{\pi}{7}[\cos\frac{6\pi}{7}+ \cos\frac{2\pi}{7}] =-1 $

$4\cos\frac{\pi}{7}[\cos\frac{2\pi}{7}-\cos\frac{\pi}{7}] =-1 $

$4\cos\frac{\pi}{7}\cdot\cos\frac{2\pi}{7} - 4\cos^{2}\frac{\pi}{7} = -1$

Further simplification will lead to equality 2

$2(\cos\frac{\pi}{9}-\cos\frac{\pi}{9} + \cos\frac{3\pi}{9}- \cos\frac{4\pi}{9}) = 1$

by using transformation formula we can prove above one also

But when it comes to following equalities

$2(\cos\frac{\pi}{11}-\cos\frac{2\pi}{11} + \cos\frac{3\pi}{11}- \cos\frac{4\pi}{11}+ \cos\frac{5\pi}{11} )= 1$

$2(\cos\frac{\pi}{13}-\cos\frac{2\pi}{13} + \cos\frac{3\pi}{13}- \cos\frac{4\pi}{13}+ \cos\frac{5\pi}{13} -\cos\frac{6\pi}{13} )= 1$

I was able to check the results with brute force in Wolfram|Alpha for above equalities, but not able to get the steps properly even though I tried manually

My question is how to generalise the summation formula and is there any method other than trigonometric approach? $$2\sum_{k=1}^n{(-1)^{k-1} \cos\frac{k\pi}{2n+1}} = 1$$

ACB
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Sivakumar Krishnamoorthi
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    Did you try to use complex numbers? $\cos \frac{\pi}{2n+1}$ is the real part of a $4n+2$-th root of unity $\xi$, and $\cos \frac{k\pi}{2n+1}$ is $Re(\xi^k)$. – Zeo Feb 23 '22 at 14:21
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    Hint take the real part of $\sum_{k}(-1)^{k-1}e^{\frac{ik}{2n+1}}$... – Surb Feb 23 '22 at 14:21
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    Your sum is $-2\Re\sum_{k=1}^nz^k$ with $z=-e^{i\pi/(2n+1)}$, i.e.$$2\Re\left[\frac{(-e^{i\pi/(2n+1)})^{n+1}+e^{i\pi/(2n+1)}}{1+e^{i\pi/(2n+1)}}\right],$$which I leave you to evaluate. – J.G. Feb 23 '22 at 14:32
  • Or use $(-1)^{k-1}\cos\varphi=\cos\big(\varphi+(k-1)\pi\big)$ and this. – metamorphy Feb 23 '22 at 15:06
  • With Mathematica : FullSimplify[ Sum[2*(-1)^(k - 1)*Cos[(k Pi)/(2 n + 1)], {k, 1, n}] == 1, Assumptions -> {n \[Element] PositiveIntegers}] gives True. – Mariusz Iwaniuk Feb 23 '22 at 16:17

2 Answers2

2

Let's use the identity $(-1)^a\cos x=\cos(a\pi-x)$. Therefore,

$$\begin{align}&2\sum_{k=1}^n{(-1)^{k-1} \cos\frac{k\pi}{2n+1}} \\= &2\sum_{k=1}^n\cos\left((k-1)\pi-\frac{k\pi}{2n+1}\right)\\=&-2\sum_{k=1}^n\cos\frac{2nk\pi}{2n+1}\end{align}$$

Take a factor $\csc\left(\frac{n\pi}{2n+1}\right)$ and then use product to sum formula.

$$\begin{align}&-\csc\left(\frac{n\pi}{2n+1}\right)\sum_{k=1}^n2\cos\frac{2nk\pi}{2n+1}\sin\frac{n\pi}{2n+1}\\=&-\csc\frac{n\pi}{2n+1}\sum_{k=1}^n\left[\sin\left((2k+1)\frac{n\pi}{2n+1}\right)-\sin\left((2k-1)\frac{n\pi}{2n+1}\right)\right]\end{align}$$

This is a telescoping sum and we are left with, $$\csc\left(\frac{n\pi}{2n+1}\right)\left[\sin\left(\frac{n\pi}{2n+1}\right)-\sin(n\pi)\right]$$

which is clearly equal to $1$.

ACB
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1

Let $z=e^{\frac{\pi i}{2 n+1}}$ , then $$ \begin{aligned} \sum_{k=1}^{n}(-1)^{k-1} z^{k} &=\frac{z\left[1-(-z)^{n}\right]}{1+z} \\ &=\frac{e^{\frac{\pi i}{2 n+1}}\left(1+(-1)^{n+1} e^{\frac{n \pi i}{2 n+1}}\right)}{1+e^{\frac{\pi i}{2 n+1}}} \end{aligned} $$

Multiplying both the numerator and denominator by $e^{-\frac{\pi i}{2(2 n+1)}}$ yields $$ \begin{aligned}\sum_{k=1}^{n}(-1)^{k-1} z^{k}& = \frac{e^{\frac{\pi i}{2(2 n+1)}}+(-1) ^{n+1} e^{\left(\frac{\pi i}{2(2 n+1)}+\frac{n \pi i}{2 n+1}\right)}}{e^{\frac{\pi i}{2(2 n+1)}}+e^{-\frac{\pi i}{2(2 n+1)}}}\\ &= \frac{e^{\frac{\pi i}{2(2 n+1)}}+(-1)^{n+1} e^{\frac{\pi}{2} i}}{e^{\frac{\pi i}{2(2 n+1)}+} e^{-\frac{\pi i}{2(2 n+1)}}}\\&= \frac{\cos \left(\frac{\pi}{2(2 n+1)}\right)+i \sin \left(\frac{\pi}{2(2 n+1)}\right)+(-1)^{n+1} i}{2 \cos \left(\frac{\pi}{2(2 n+1)}\right)} \qquad\qquad (*) \end{aligned} $$

Comparing the real parts of (*) yields $$ \sum_{k=1}^{n}(-1)^{k-1} \cos \left(\frac{k \pi}{2 n+1}\right)=\frac{1}{2} $$

By the way, comparing the imaginary parts of (*) gives $$ \sum_{k=1}^{n}(-1)^{k-1} \sin \left(\frac{k \pi}{2 n+1}\right)=\frac{1}{2}\left[\tan \left(\frac{\pi}{2(2 n+1)}\right)+(-1)^{n+1} \sec \left(\frac{\pi}{2(2 n+1)}\right)\right] $$

Lai
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