## Theorem

Let $p$ be a positive odd prime.

Then:

$\forall m, n \in \N: \displaystyle \size {\sum_{k \mathop = m}^{m + n} \paren {\frac k p} } < \sqrt p \, \ln p$

where $\paren {\dfrac k p}$ is the Legendre symbol.

## Proof

 $\displaystyle \sum_{k \mathop = m}^{m + n} \paren {\dfrac k p}$ $=$ $\displaystyle \frac 1 p \sum_{k \mathop = 0}^{p - 1} \sum_{x \mathop = m}^{m + n} \sum_{a \mathop = 0}^{p - 1} \paren {\dfrac k p} e^{2 \pi i a \paren {x - k} / p}$ $\displaystyle$ $=$ $\displaystyle \frac 1 p \sum_{a \mathop = 1}^{p - 1} \sum_{x \mathop = m}^{m + n} e^{2 \pi i a x / p} \sum_{k \mathop = 0}^{p - 1} \paren {\dfrac k p} e^{-2 \pi i a t / p}$

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 $\sqrt p$.

Hence:

 $\displaystyle \size {\sum_{t \mathop = m}^{m + n} \paren {\dfrac t p} }$ $\le$ $\displaystyle \size {\frac {\sqrt p} p \sum_{a \mathop = 1}^{p - 1} \sum_{x \mathop = m}^{m + n} e^{2 \pi a i x / p} }$ $\displaystyle$ $=$ $\displaystyle \size {\frac {\sqrt p} p \sum_{a \mathop = 1}^{p - 1} e^{2 \pi i a m / p} \sum_{x \mathop = 0}^n e^{2 \pi i a x / p} }$ $\displaystyle$ $\le$ $\displaystyle \size {\frac {\sqrt p} p \sum_{a \mathop = 1}^{p - 1} \frac {e^{2 \pi i a n / p} - 1} {e^{2 \pi i a / p} - 1} }$ $\displaystyle$ $=$ $\displaystyle \size {\frac {\sqrt p} p \sum_{a \mathop = 1}^{p - 1} \frac {e^{\pi i a n / p} \, \map \sin {\pi a n / p} } {e^{\pi i a / p} \, \map \sin {\pi a / p} } }$ $\displaystyle$ $\le$ $\displaystyle \frac {\sqrt p} p \sum_{a \mathop = 1}^{p - 1} \size {\frac 1 {\map \sin {\pi \left\langle{a / p}\right\rangle} } }$ $\displaystyle$ $\le$ $\displaystyle \frac {\sqrt p} p \sum_{a \mathop = 1}^{p - 1} \frac 1 {2 \norm {a / p} }$

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:

 $\displaystyle \frac 1 2 \sum_{a \mathop = 1}^{p-1} \frac 1 {\norm{a / p} }$ $=$ $\displaystyle \sum_{0 \mathop < a \mathop < \frac p 2} \frac p a$ $\displaystyle$ $=$ $\displaystyle p \sum_{a \mathop = 1}^{\frac {p - 1} 2} \frac 1 a$

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:

 $\displaystyle \size {\sum_{t \mathop = m}^{m + n} \paren {\frac t p} }$ $\le$ $\displaystyle \frac {\sqrt p} p \cdot p \sum_{a \mathop = 1}^{\frac {p - 1} 2} \frac 1 a$ $\displaystyle$ $<$ $\displaystyle \sqrt p \sum_{a \mathop = 1}^{\frac {p - 1} 2} \ln \frac {2 a + 1} {2 a - 1}$ $\displaystyle$ $=$ $\displaystyle \sqrt p \, \ln p$

$\blacksquare$

## Source of Name

This entry was named for George Pólya and Ivan Matveevich Vinogradov.