Start with the following manipulations:

The expression $\sum_{{t=0}}^{{p-1}}(\frac{t}{p})e^{{-2\pi iat/p}}$ is just a Gauss sum, and has magnitude $\sqrt{p}\,$. Hence

Here $\langle x\rangle$ denotes the absolute value of the difference between $x$ and the closest integer to $x$, i.e. $\langle x\rangle=\inf_{{z\in{\bf Z}}}\{|x-z|\}$.

Since $p$ is odd, we have

Now $\ln\frac{2x+1}{2x-1}>\frac{1}{x}$ for $x>1$; to prove this, it suffices to show that the function $f:[1,\infty)\rightarrow{\bf R}$ given by $f(x)=x\ln\frac{2x+1}{2x-1}$ is decreasing and approaches 1 as $x\rightarrow\infty$. To prove the latter statement, substitute $v=1/x$ and take the limit as $v\rightarrow 0$ using L’Hôpital’s rule. To prove the former statement, it will suffice to show that $f^{{\prime}}$ is less than zero on the interval $[1,\infty)$. But $f^{{\prime}}(x)\rightarrow 0$ as $x\rightarrow\infty$ and $f^{{\prime}}$ is increasing on $[1,\infty)$, since $f^{{\prime\prime}}(x)=\frac{-4}{4x^{2}-1}(1-\frac{4x^{2}+1}{4x^{2}-1})>0$ for $x>1$, so $f^{{\prime}}$ is less than zero for $x>1$.

With this in hand, we have

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