Trigonometric sums and limits

Using de Moivre, we can find that

$S(n)=Im(e^{ix}+...+e^{inx})$

After a bit of work with the geometric series formula we get that

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

Then we sum S(n) from 1 to p:

$\displaystyle \sum_{n=1}^{p}S(n)=\sum_{n=1}^{p}\frac{\cos({x/2})-\cos((n+1/2)x)}{2\sin(x/2)} = p.\frac{\cos({x/2})}{2\sin(x/2)}-\frac{1}{2\sin(x/2)}.\sum_{n=1}^{p} {\cos((n+1/2)x)}$

Similar to the way we figured out S(n), we can find a formula for the sum of cosines, namely

$\frac{\sin(x)+\sin(x(p+1))}{2\sin(x/2)}$

But in the question we are dealing with 1/p.(the sum) as p goes to infinity.Then everything will go to zero (sine is a bounded function), thus we are left with only

$\frac{1}{p}.p.\frac{\cos({x/2})}{2\sin(x/2)}$

Canceling the p's and plugging in x=pi/3, we get the desired answer

Let us calculate $\lim_{n\to\infty} S(n)$ if it exists then by Cauchy's First Limit Theorem the limit of Arithmetic Mean of $S(n)$ as n tends to infinity is equal to the limit of $S(n)$ . Note also that $S(n) = \sum_{k=0}^{\infty} sin(kx)$

$\lim_{n\to\infty} S(n) = \Im (\frac{1}{1-e^{\frac{i\pi}{3}}})$

Note that I have used the Taylor Expansion for $\frac{1}{1-z} = \sum_{r=0}^{\infty} z^{r}$ which is valid for $\lvert(z)\rvert \leq 1$ (except for $z=1$ )

So here $z = e^{\frac{i\pi}{3}}$ .

So our answer is $\Im (\frac{1}{1-\cos(\frac{\pi}{3}) - i\sin(\frac{\pi}{3})})$

$=\sin(\frac{\pi}{3})$

= $\frac{\sqrt{3}}{2}$

Here $\Im(.)$ denotes the imaginary part of complex Number