# Summation of sin

Hello,
Can anyone give some hints on how to solve this:

$$\sum_{n=0}^{K-1}\frac{sin(2\pi n^2\Delta)}{n}$$

It's just the n^2 that complicates things. I tried re-writing it as

$$Im\sum_{n=0}^{K-1}\frac{e^{j n^2 x}}{n}$$,

where $$x=2\pi \Delta$$
but I cannot solve this either.

Thanks,
svensl

Last edited:

What is delta? If it is an integer than sin(2*pi*k) for any integer k is equal to 0.