A series in cosine : $\sum_{j=1}^{n} (-1)^{j-1}\cos\left(\frac{j\pi}{2n+1}\right)$

Question

$\textbf{Question :}$ Prove that $$\sum_{j=1}^{n} (-1)^{j-1}\cos\left(\frac{j\pi}{2n+1}\right)=\frac{1}{2}$$

$\textbf{My Attempt :}$

Let $n$ be odd (will do even case later) then the series can be written as

$$S = \sum_{j=1}^{(n+1)/2}\cos((2j-1)x) - \sum_{j=1}^{(n-1)/2} \cos(2jx)$$

where $x = \frac{\pi}{2n+1}$

Now as both the cosine series are just arithmetic progressions in their angles with the common difference of $2x$, we can simplify $S$ as follows :

$$S = \frac{1}{\sin x}(\sin((n+3)x)\cos((n+2)x)-\sin((n+1)x)\cos(nx))$$

As $2\sin a\cos b = \sin(a+b)+\sin(a-b)$

$$S = \frac{1}{2\sin x}(\sin(2n+5)x)-\sin((2n+1)x)$$

Again as $\sin c -\sin d = 2\sin((c+d)/2)\cos((c-d)/2)$

$$S = \frac{1}{2 \sin x} (2\sin 2x \cos((2n+3)x))$$

$$S = \cos x \cos ((2n+3)x)$$ Now this is where I am getting stuck, I can't get it equal to any constant...

Answer 1

We have \begin{align*} &\cos \frac{x}{2} \cdot \sum_{j=1}^{n} (-1)^{j-1}\cos(j x)\\ ={}& \sum_{j=1}^n (-1)^{j-1} \frac12[\cos(j x - x/2) + \cos(jx + x/2)]\\ ={}& \sum_{j=1}^n (-1)^{j-1} \frac12\cos(j x - x/2) + \sum_{j=1}^n (-1)^{j-1} \frac12\cos(jx + x/2)\\ ={}&\frac12 \cos \frac{x}{2} + \sum_{j=2}^n (-1)^{j-1} \frac12\cos(j x - x/2) + \sum_{j=1}^{n-1} (-1)^{j-1} \frac12\cos(jx + x/2) \\ &\qquad + (-1)^{n-1}\frac12\cos(nx + x/2)\\ ={}&\frac12 \cos \frac{x}{2} - \sum_{j=1}^{n-1} (-1)^{j-1} \frac12\cos(j x + x/2) + \sum_{j=1}^{n-1} (-1)^{j-1} \frac12\cos(jx + x/2)\\ &\qquad + (-1)^{n-1}\frac12\cos(nx + x/2)\\ ={}& \frac12 \cos \frac{x}{2} + (-1)^{n-1}\frac12\cos(nx + x/2) \end{align*} which results in (when $\cos (x/2) \ne 0$) $$\sum_{j=1}^{n} (-1)^{j-1}\cos(j x) = \frac12 + \frac{(-1)^{n-1}\cos(nx + x/2)}{2\cos (x/2)}.$$

Letting $x = \frac{\pi}{2n + 1}$, the desired result follows.

Answer 2

Using $-\cos(x) = \cos(\pi -x)$ we can write $$ S =\sum_{j=1}^{n} (-1)^{j-1}\cos\left(\frac{j\pi}{2n+1}\right) $$ in the form $\sum_{k=0}^{n-1}\cos (a+k \cdot d)$, which is a well-known expression.

Let $N$ be the largest odd integer $\le n$, then $2n-1-N$ is the largest even integer $\le n$. The terms for odd $j$ in $S$ are $$ \cos\left(\frac{\pi}{2n+1}\right) + \cos\left(\frac{3\pi}{2n+1}\right)+ \cdots + \cos\left(\frac{N\pi}{2n+1}\right) $$ and the terms for even $j$ in $S$ are $$ -\cos\left(\frac{2\pi}{2n+1}\right) - \cos\left(\frac{4\pi}{2n+1}\right)- \cdots - \cos\left(\frac{(2n-1-N)\pi}{2n+1}\right)\\ = \cos\left(\frac{(2n-1)\pi}{2n+1}\right) + \cos\left(\frac{(2n-3)\pi}{2n+1}\right)+ \cdots + \cos\left(\frac{(N+2)\pi}{2n+1}\right) \, . $$ Therefore, regardless whether $n$ is even or odd, the given sum is equal to $$ S = \sum_{k=0}^{n-1} \cos\left(\frac{(2k+1)\pi}{2n+1}\right) \, . $$ That sum can be evaluated with well-known techniques, see for example How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?: $$ \sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \cos \biggl( \frac{ 2 a + (n-1)\cdot d}{2}\biggr) \, . $$

Here we have $a = \pi/(2n+1)$ and $d = 2\pi/(2n+1)$, so that $$ S = \frac{\sin\left(\frac{n}{2n+1}\pi\right)\cos \left(\frac{n}{2n+1}\pi\right)}{\sin\left(\frac{1}{2n+1}\pi\right)} = \frac{\sin\left(\frac{2n}{2n+1}\pi\right)}{2 \sin\left(\frac{1}{2n+1}\pi\right)} = \frac 12 \, . $$