## Loser66 one year ago Problem is in comment

1. Loser66

2. Loser66

There is an "i" on the Gauss sums. It's a typo.

3. anonymous

Let me just retype the problem here, MSWord always gives me trouble. $\sum_{n=1}^N\exp\frac{2i\pi n^2}{N}=\begin{cases} \sqrt N&\text{if }N\equiv1\mod4\\ 0&\text{if }N\equiv 2\mod4\\ i\sqrt N&\text{if }N\equiv3\mod4\\ (1+i)\sqrt N&\text{if }N\equiv0\mod4 \end{cases}$ and you want to prove that $$\sum\limits_{n\ge1}\dfrac{z^{n^2}}{n}$$ converges for $$z=\exp\dfrac{2i\pi}{N}$$ if $$N\equiv2\mod4$$ and diverges otherwise.

4. anonymous

Did I get everything right?

5. Loser66

Yes, it is right. My prof said that we need use Dirichlet test to attack it. From part 1, we have $$\sum_{n\geq 1} \dfrac{z^{n^2}}{n}$$ converges with radius R =1.

6. anonymous

For the case $$N\equiv2\mod4$$, you have $$z=\exp\dfrac{2i\pi}{N}$$ and $\sum_{n=1}^N\frac{z^{n^2}}{n}=\sum_{n=1}^N\frac{\exp\dfrac{2i\pi n^2}{N}}{n}$ Call $$a_n=\dfrac{1}{n}$$ and $$b_n=\exp\dfrac{2i\pi n^2}{N}$$. To satisfy Dirichlet's test, you must (1) $$a_n$$ a non-increasing and converging to $$0$$, and (2) $$\left|\sum\limits_{n=1}^Nb_n\right|\le M$$ for some constant $$M$$ and all $$N$$. The first condition is clearly satisfied. Meanwhile, $\left|\sum_{n=1}^N\exp\dfrac{2i\pi n^2}{N}\right|=\left|\sum_{n=1}^N0\right|=0\le M$

7. anonymous

... and so $$\sum a_nb_n$$ converges for that case.

8. Loser66

I got it,

9. Loser66

Question: the sum we need prove is infinite sum, not stop at N. How to argue that part?

10. anonymous

Slight typo above: In that last line, it should just be $$0$$, not the sum of $$0$$s. The procedure is similar for the other case. Set $$N\equiv1\mod4$$, use the same $$a_n$$, and replace $$b_n$$ accordingly. You have $\left|\sum_{n=1}^N\exp\dfrac{2i\pi n^2}{N}\right|=\left|\sqrt N\right|=\sqrt N$ which is unbounded, so the second required condition is not met, and the series diverges.

11. anonymous

The test is designed to check the convergence of the series $$\sum\limits_{n=1}^\infty a_nb_n$$, while one of the conditions for the test is to check the boundedness of a partial sum, $$\sum\limits_{n=1}^Nb_n$$.

12. anonymous

The case for $$N\equiv3\mod4$$ is almost identical as for $$N\equiv 1\mod4$$. The partial sum of $$b_n$$ is equal to $$|i\sqrt N|=\sqrt N$$, and you get the same conclusion. For $$N\equiv0\mod4$$, you have $$|(1+i)\sqrt N|=\sqrt{2N}$$, etc.

13. Loser66

Thank you so so so much. You make it simple.

14. anonymous

You're welcome!

15. Loser66

oh, I got it, nvm. :)