For a time series with polynomial growth, auto-correlation $\hat\rho_n(k) \longrightarrow 1$ as $n \longrightarrow \infty$ I'm trying to do the following exercise:

Let $(x_i)_{1 \le i \le n}$ be a time series. Then the empirical auto-correlation function $\hat\rho: \{1,\ldots,n-1\} \to \mathbb R$ is defined by $$\hat\rho_n(k) = \frac{\sum_{i=1}^{n-k}(x_i -\bar x)(x_{i+k}-\bar x)}{\sum_{i=1}^{n} (x_i-\bar x)^2}$$
Fix $d \in \mathbb N_{>0}$ and $(a_0,\ldots,a_d) \in \mathbb R^{d+1}$. Now consider a particular time series in which $x_i = \sum_{j=0}^d a_j i^j$ for all $1 \le i \le n$. Prove that

For a fixed $k$, $\hat\rho_n(k) \longrightarrow 1$ as $n \longrightarrow \infty$.


I try to plug $$\bar x = \frac{1}{n} \sum_{j=0}^d a_j \sum_{i=1}^n i^j$$
in $(x_i -\bar x)(x_{i+k}-\bar x)$ and get $$\begin{aligned}(x_i -\bar x)(x_{i+k}-\bar x) &= \left (\sum_{j=0}^d a_j i^j -\frac{1}{n} \sum_{j=0}^d a_j \sum_{i=1}^n i^j \right) \left(\sum_{j=0}^d a_j (i+k)^j-\frac{1}{n} \sum_{j=0}^d a_j \sum_{i=1}^n i^j \right) \\
&= \left (\sum_{j=0}^d a_j \left ( i^j -\frac{1}{n}  \sum_{i=1}^n i^j \right) \right) \left(\sum_{j=0}^d a_j \left ( (i+k)^j-\frac{1}{n}  \sum_{i=1}^n i^j \right ) \right)\end{aligned}$$
Then I'm stuck at simplifying this expression and can not move on. Could you please elaborate on how to solve this problem?
 A: The proof exploits the fact that the growth in $(x_i)$ is polynomial. To get $\hat\rho_n$ to look like $1$, the obvious step is to write
$$(x_i-\bar x)(x_{i+k}-\bar x) = (x_i-\bar x)^2 + (x_i-\bar x)(x_{i+k}-x_i)
$$
so that
$$
\hat\rho_n=1-\frac{\sum_{i=n-k+1}^n(x_i-\bar x)^2}{\sum_1^n(x_i-\bar x)^2}+\frac{\sum_{i=1}^{n-k}(x_i-\bar x)(x_{i+k}-x_i)}{\sum_1^n(x_i-\bar x)^2}=:1-\varepsilon_n+\delta_n.
$$
The aim is to show that both  $\varepsilon_n$ and $\delta_n$ tend to zero.  To do this, recall that $x_i= f(i)$ where $f$ is the polynomial function
$$f(t):=\sum_{j=0}^da_j t^j.$$
We can assume wlog that the coefficient $a_d$ on the highest power $t^d$ is nonzero. It follows that, as $n\to\infty$:
$$\begin{aligned}
& x_i\sim a_d n^d\quad\text{if $i>n-k$}\\
&\sum_1^n x_i\sim \frac{a_d}{d+1}n^{d+1}\\
&\sum_1^n x_i^2\sim \frac{a_d^2}{2d+1}n^{2d+1}\\\end{aligned}$$
which in turn imply (as $n\to\infty$)
$$\begin{aligned}
&\bar x\sim\frac{a_d}{d+1}n^d\\
&x_i - \bar x\sim\frac {da_d}{d+1}n^d\qquad\text{if $i>n-k$}\\
&\sum_1^n(x_i-\bar x)^2=\sum_1^nx_i^2-n\bar x^2\sim \frac{a_d^2}{2d+1}n^{2d+1}-\left(\frac{a_d}{d+1}\right)^2n^{2d+1}
\end{aligned}
$$
and therefore $\varepsilon_n\to0$. To handle $\delta_n$,  write
$$
\delta_n^2\le \frac{\sum_{i=1}^{n-k}(x_i-\bar x)^2\sum_{i=1}^{n-k}(x_{i+k}-x_i)^2}{[\sum_1^n(x_i-\bar x)^2]^2}\le\frac{\sum_{i=1}^{n-k}(x_{i+k}-x_i)^2}{\sum_1^n(x_i-\bar x)^2}
$$
where the first inequality is Cauchy-Schwarz. Bound the numerator of the RHS using the mean value theorem:
$$|x_{i+k}-x_i|=|f(i+k)-f(i)|=|f'(i^*)k|,
$$
where $i^*$ is between $i$ and $i+k$. Since $f'$ is a polynomial of degree $d-1$, the RHS of the above is at most $A+Bn^{d-1}$ for  positive constants $A$ and $B$, so that $\sum_{i=1}^{n-k}(x_{i+k}-x_i)^2=O(n^{2d-1})$. Since   $\sum_1^n(x_i-\bar x)^2=\Theta(n^{2d+1})$, conclude  $\delta_n\to0$.
