Let $p$ be a prime number, and $\zeta = e^{(2\pi i/p)}$ be a $p$th root of unity. Given the gauss sum: $$g_t = \sum_{k = 0}^{p-1}\left(\frac{k}{p}\right)\zeta^{tk}$$

I'm trying to prove the upper bound:

$$\left|\sum_{t = m}^n g_t\right|< p\log p$$

for arbitrary $m$ and $n$. This will be straightforward if I manage to find bounds for:

$$\left|\sum_{k = m}^n \zeta^{tk}\right|$$

when $t \neq 0 \pmod p$, especially if this bound is $\leq \frac{p\log p}{p-1}$. I've tried to exponentiate the sum and turn it into a product but I'm not really sure what $e^\zeta$ looks like. Any tips?

EDIT: By just applying a sequence of triangle inequalities, I've managed to find:

$$\left| \sum_{t = m}^n g_t\right| < p(p-1)$$

So I guess that it needs a more ingenious way to lower this bound.


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