Start with the following manipulations: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = m}^{m + n} \paren {\dfrac k p}       | r = \frac 1 p \sum_{k \mathop = 0}^{p - 1} \sum_{x \mathop = m}^{m + n} \sum_{a \mathop = 0}^{p - 1} \paren {\dfrac k p} e^{2 \pi i a \paren {x - k} / p}       | c =  }} {{eqn | r = \frac 1 p \sum_{a \mathop = 1}^{p - 1} \sum_{x \mathop = m}^{m + n} e^{2 \pi i a x / p} \sum_{k \mathop = 0}^{p - 1} \paren {\dfrac k p} e^{-2 \pi i a t / p}        | c =  }} {{end-eqn}} The expression: :$\displaystyle \sum_{k \mathop = 0}^{p - 1} \paren {\dfrac k p} e^{-2 \pi i a t / p}$ is a Gauss sum, and has magnitude $\sqrt p$. {{explain|The above statement.}} Hence: {{begin-eqn}} {{eqn | l = \size {\sum_{t \mathop = m}^{m + n} \paren {\dfrac t p} }       | o = \le       | r = \size {\frac {\sqrt p} p \sum_{a \mathop = 1}^{p - 1} \sum_{x \mathop = m}^{m + n} e^{2 \pi a i x / p} }       | c =  }} {{eqn | r = \size {\frac {\sqrt p} p \sum_{a \mathop = 1}^{p - 1} e^{2 \pi i a m / p} \sum_{x \mathop = 0}^n e^{2 \pi i a x / p} }       | c =  }} {{eqn | o = \le       | r = \size {\frac {\sqrt p} p \sum_{a \mathop = 1}^{p - 1} \frac {e^{2 \pi i a n / p} - 1} {e^{2 \pi i a / p} - 1} }       | c =  }} {{eqn | r = \size {\frac {\sqrt p} p \sum_{a \mathop = 1}^{p - 1} \frac {e^{\pi i a n / p} \, \map \sin {\pi a n / p} } {e^{\pi i a / p} \, \map \sin {\pi a / p} } }       | c =  }} {{eqn | o = \le       | r = \frac {\sqrt p} p \sum_{a \mathop = 1}^{p - 1} \size {\frac 1 {\map \sin {\pi \left\langle{a / p}\right\rangle} } }       | c =  }} {{eqn | o = \le       | r = \frac {\sqrt p} p \sum_{a \mathop = 1}^{p - 1} \frac 1 {2 \norm {a / p} }       | c =  }} {{end-eqn}} {{explain|$\left\langle{a / p}\right\rangle$}} Here $\norm x$ denotes the difference between $x$ and the closest integer to $x$, that is: :$\displaystyle \norm x = \inf_{z \mathop \in \Z} \set {\size {x - z} }$ Since $p$ is odd, we have: {{begin-eqn}} {{eqn | l = \frac 1 2 \sum_{a \mathop = 1}^{p-1} \frac 1 {\norm{a / p} }       | r = \sum_{0 \mathop < a \mathop < \frac p 2} \frac p a       | c =  }} {{eqn | r = p \sum_{a \mathop = 1}^{\frac {p - 1} 2} \frac 1 a       | c =  }} {{end-eqn}} Now: :$\ln \dfrac {2 x + 1} {2 x - 1} > \dfrac 1 x$ for $x > 1$. To prove this, it suffices to show that the function $f: \hointr 1 \infty \to \R$ given by: :$\map f x = x \ln \dfrac {2 x + 1} {2 x - 1}$ is decreasing and approaches $1$ as $x \to \infty$. To prove the latter statement, substitute $v = 1/x$ and take the limit as $v \to 0$ using L'Hospital's Rule. To prove the former statement, it will suffice to show that $f'$ is less than zero on the interval $\hointr 1 \infty$. Now we have that: :$\map {f''} x = \dfrac {-4} {4 x^2 - 1} \paren {1 - \dfrac {4 x^2 + 1} {4 x^2 - 1} } > 0$ for $x > 1$. So $f'$ is increasing on $\hointr 1 \infty$. But $\map {f'} x \to 0$ as $x \to \infty$. So $f'$ is less than zero for $x > 1$. With this in hand, we have: {{begin-eqn}} {{eqn | l = \size {\sum_{t \mathop = m}^{m + n} \paren {\frac t p} }       | o = \le       | r = \frac {\sqrt p} p \cdot p \sum_{a \mathop = 1}^{\frac {p - 1} 2} \frac 1 a       | c =  }} {{eqn | o = <       | r = \sqrt p \sum_{a \mathop = 1}^{\frac {p - 1} 2} \ln \frac {2 a + 1} {2 a - 1}       | c =  }} {{eqn | r = \sqrt p \, \ln p       | c =  }} {{end-eqn}} {{qed}} {{Namedfor|George Pólya|name2 = Ivan Matveevich Vinogradov|cat = Pólya|cat2 = Vinogradov I M}}