This problem is from Stein, Fourier Analysis,Chapter 4,problem 2(d).
Problem:Suppose that $P(x)=c_n x^n+……+c_0$ is a polynomial with real coefficients, where at least one of $c_1,……,c_n$ is irrational. Then the sequence $<P(n)>$ is equidistributed in $[0,1)$.
The key point is to show that $$ \lim_{N \to \infty}\ \frac{1}{N} \sum_{n=1}^N e^{2\pi i k P(n)} \ =0 $$ whenever $c_2,……,c_n$ is rational and $c_1$ irrational.
Since $<c_1n+c_0>$ is equidistributed in $[0,1)$ if $c_1$ irrational. If the equation above holds, then by induction:
Assume that polynomial $P$ is of degree $m$ with at least one of $c_1,……,c_n$ irrational satisfies that $<P(n)>$ equidistributed in $[0,1)$, now consider the condition of degree $m+1$.We may discuss on two conditions.
(i)only $c_1$ is irrational, then by the equations above, we immediately get the solution;
(ii)otherwise consider $P(n+h)-P(n)$ for any positive integer $h$, a polynomial of degree $m$, along with at lease one irrational coefficient. By assumption, $<P(n+h)-P(n)>$ is equidistributed in $[0,1)$,and hence $<P(n)>$ is equidistributed in $[0,1)$.
My only drawback is to estimate the sum $$ \sum_{n=1}^N e^{2 \pi i k P(n)} $$ to conclude that $$ \lim_{N \to \infty}\ \frac{1}{N} \sum_{n=1}^N e^{2\pi i k P(n)} \ =0 $$
Any help will be appreciated, thank you.
After a simple classification I have figured it out.
We could solve it by induction on the degree of the polynomial $P(x)$.
If $deg P(x)=1$, then $c_1 \in \mathbb{R} \setminus \mathbb{Q}$ and apparently $<c_1n+c_0>$ is equidistributed in $[0,1)$.
Assume $deg P(x)=m$ and at least one of its coefficient $c_1,……,c_m$ is irrational, then $<P(n)>$ is equidistributed in $[0,1)$.
Now consider the situation of $deg P(x)=m+1$, we may divide it into two parts:
(i)If $c_{m+1} \in \mathbb{R} \setminus \mathbb{Q}$, then for any $h \in \Bbb{Z_+}$ we have $$P(n+h)-P(n)=\sum_{j=0}^m \bar{c_j}x^j$$ a polynomial of degree $m$, along with $\bar{c_m}$ irrational. Then by the assumption above: $<P(n+h)-P(n)>$ is equidistributed in $[0,1)$, therefore by Weyl's theorem: $<P(n)>$ is equidistributed in $[0,1)$.
(ii)If $c_{m+1}=\frac{p}{q} \in \mathbb{Q}$, then for $N$ big eough, there exists the unique $l,r$ such that $N=lq+r(0 \le r \le q-1)$. Set $P(x)=\frac{p}{q}x^{m+1} + Q(x)$, where $Q(x)$ is a polynomial of degree $m$ with at least one of $c_1,……,c_m$ irrational.
Notice that $$ \frac{1}{N} \sum_{n=1}^N e^{2 \pi ik \frac{p}{q} \left( x^{m+1}+Q(n) \right)} =\frac{1}{N} \left[ \sum_{j=0}^l e^{2 \pi ik Q(jq)} + …… +e^{2\pi ik \frac{r^m p}{q}} \sum_{j=0}^le^{2 \pi ikQ(jq+r)} +e^{2\pi ik \frac{(r+1)^mp}{q}} \sum_{j=0}^{l-1}e^{2 \pi ik Q(jq+r+1)}+ ……+ e^{2 \pi ik \frac{(q-1)^m p}{q}}\sum_{j=0}^{l-1} e^{2 \pi ik Q(j+q-1)} \right] $$
Since $0 < \frac{l}{M} < \frac{l+1}{N} < \frac{2}{q}$, then $$\frac{1}{N} \sum_{n=1}^N e^{2 \pi ik \left( \frac{p}{q}x^{m+1}+Q(n) \right)} \to 0(N \to \infty)$$
Consequently $<P(n)>$ is equidistributed in $[0,1)$.