For a fixed $k$, $\hat\rho_n(k) \longrightarrow 1$ as $n \longrightarrow \infty$ I posted this question on MSE and have not found any answer, so I cross-post it here. I will notify when one of them gets an answer.

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 $$\hat\rho_n(k) = \frac{\sum_{i=1}^{n-k}(x_i -\bar x)(x_{i+k}-\bar x)}{\sum_{i=1}^{n} (x_i-\bar x)^2}$$ where $\bar x = (1/n) \sum_{i=1}^{n} x_i$.
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: Partial answer for $j=1$ and $k=1$ with, additionally, assuming that $a_0=0$ and hence also not fitting a constant in an AR(1) model (recall that the $k=1$st autocorrelation coefficient equals the estimated AR(1) coefficient):
So we consider $y_t=\delta t+ \epsilon_t$, fit an $AR(1)$ model without constant to $y_t$, and show that $\hat{\rho}\to_p1$.
\begin{eqnarray*}
  \hat{\rho}&=&\frac{\sum_ty_{t-1}y_{t}}{\sum_ty_{t-1}^2}\\
  &=&\frac{\sum_t(\delta(t-1)+\epsilon_{t-1})(\delta t+\epsilon_{t})}{\sum_t(\delta(t-1)+\epsilon_{t-1})^2}\\
  &=&\frac{\sum_t\delta^2(t-1)t+\delta(t-1)\epsilon_{t}+\delta t\epsilon_{t-1}+\epsilon_{t-1}\epsilon_{t}}{\sum_t\delta^2(t-1)^2+2\delta(t-1)\epsilon_{t-1}+\epsilon_{t-1}^2}\\
  &=&\frac{\frac{1}{T^3}\sum_t\delta^2t^2-\delta^2t+\delta(t-1)\epsilon_{t}+\delta t\epsilon_{t-1}+\epsilon_{t-1}\epsilon_{t}}{\frac{1}{T^3}\sum_t\delta^2(t^2-2t+1)+2\delta(t-1)\epsilon_{t-1}+\epsilon_{t-1}^2}\\
  &=&\frac{\frac{1}{T^3}\sum_t\delta^2t^2-\delta^2t+\delta(t-1)\epsilon_{t}+\delta t\epsilon_{t-1}+\epsilon_{t-1}\epsilon_{t}}{\frac{1}{T^3}\sum_t\delta^2(t^2-2t+1)+2\delta(t-1)\epsilon_{t-1}+\epsilon_{t-1}^2}\\
  &=&\frac{\frac{\delta^2}{3}}{\frac{\delta^2}{3}}+o_p(1)\\
  &=&1+o_p(1),
  \end{eqnarray*}
where the second-to-last line follows because all other sums vanish when divided by $T^3$. For $\delta(t-1)\epsilon_{t}$ see e.g. Hamilton, Prop. 17.1(c).
