# Value of given series $\sum_{n=1}^{n=\infty}\frac{\cos{\frac{n\pi}{3}}}{n+1}$ and $\sum_{n=1}^{n=\infty}\frac{\sin{\frac{n\pi}{3}}}{n+1}$

I was doing a problem,after some calculation it all came down to the value of series $$\sum_{n=1}^{n=\infty}\frac{\cos{\frac{n\pi}{3}}}{n+1}$$ and $$\sum_{n=1}^{n=\infty}\frac{\sin{\frac{n\pi}{3}}}{n+1}$$ What is the value of this series?Can anybody help me?And also help with method to find generelised version of this as $\cos{np}$ and $\sin{np}$ where $p$ is a positive nonzero constant

• Wolfram Alpha gives $\frac{\pi} {2\sqrt{3}}-1$ for the first and $\frac {\pi} {6}$ for the second. – Cameron Williams Jan 4 '15 at 14:49

$$\sum_{n=1}^{\infty}\frac{\cos{\frac{n\pi}{3}}}{n+1}+i\sum_{n=1}^{\infty}\frac{\sin{\frac{n\pi}{3}}}{n+1}$$

$$=\sum_{n=1}^{\infty}\frac{e^\frac{i n\pi}3}{n+1} =e^\frac{-i\pi}3\sum_{n=1}^{\infty}\frac{\left(e^\frac{i\pi}3\right)^{n+1}}{n+1}$$

Now for $-1<x<1,$ $$\sum_{n=1}^{\infty}\frac{x^{n+1}}{n+1}=-\ln(1-x)-x$$

Finally, $\ln(1-e^{2iy})=\ln[e^{iy}(e^{-iy}-e^{iy})]=\ln(e^{iy})+\ln(e^{-iy}-e^{iy})$

$=iy+\ln(-2i\sin y)=\cdots$

Can you take it from here?

• Thank you,perfect answer,oh also I got the problem correctly :-) – jjoyk Jan 4 '15 at 15:08