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loser66 · Answer

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loser66 · Answer

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

anonymous · Answer

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.

anonymous · Answer

Did I get everything right?

loser66 · Answer

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.

anonymous · Answer

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,
$$\left|\sum_{n=1}^N\exp\dfrac{2i\pi n^2}{N}\right|=\left|\sum_{n=1}^N0\right|=0\le M$$

anonymous · Answer

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

loser66 · Answer

I got it,

loser66 · Answer

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

anonymous · Answer

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.

anonymous · Answer

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, $\sum\limits_{n=1}^Nb_n$.

anonymous · Answer

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.

loser66 · Answer

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

anonymous · Answer

You're welcome!

loser66 · Answer

oh, I got it, nvm. :)