Consider a sequence defined by $$ x_n =n^{-1/2}\sum_{t=1}^{\lfloor n/p\rfloor p}(e^{2\pi i/p})^{(k-l\lfloor n/p\rfloor p/n)t} $$ for $n\ge1$, where $p\ge2$, $i=\sqrt{-1}$, $k,l\in\mathbb Z$ such that $k-l\in\mathbb Z\setminus p\mathbb Z$ (i.e. $k-l$ is not an integer multiple of $p$). How can I show that $x_n\to0$ as $n\to\infty$?
When $n$ is an integer multiple of $p$, then $$ x_n =\sum_{t=1}^{n}(e^{2\pi i/p})^{(k-l)t}=0 $$ since $k-l\in\mathbb Z\setminus p\mathbb Z$. We also have that $\lfloor n/p\rfloor p/n\to1$ and $\lfloor n/p\rfloor p\sim n$ as $n\to\infty$. So it seems that $x_n\to0$ as $n\to\infty$ but I cannot find an argument that verifies that this is in deed true. How can we show that $x_n\to0$ as $n\to\infty$?
Any help is much appreciated!
Let $r$ be the remainder of the division of $n$ by $p$ and $q$ be the quotient. We can see that $$x_n=O(n^{-1/2})+n^{-1/2}\sum_{s=0}^{q-1}{\sum_{t=0}^{p-1}{e^{2i\pi t(k-l)/p}e^{2i\pi lr(ps+t)/(pn)}}}=O(n^{-1/2})+n^{-1/2}\sum_{s=0}^{q-1}{e^{2i\pi lrs/n}}\sum_{t=0}^{p-1}{e^{2i\pi [n(k-l)+lr]t/pn}}.$$
Write the first sum in the RHS as $U_n$, and the second one as $V_n$. It’s easy to see $U_n=O(n)$. So if we show $V_n=o(n^{-1/2})$, we are done. But $$|V_n|=\left|V_n-\sum_{t=0}^{p-1}{e*{2i\pi (k-l)t/p}}\right| \leq \sum_{t=0}^{p-1}{|e^{2i\pi t(k-l)/p}||e^{2i \pi lrt/pn }-1|} \leq C\sum_{t=0}^{p-1}{\frac{|l|rt}{pn}} = O(1/n),$$, hence the conclusion.
In fact, we’ve even shown that $x_n=O(n^{-1/2})$.