Prove that a specific auto-correlation function is even
Let me get into more additional detail from what Peter wrote.
First, note that $h[n]$ is the output of an IIR filter with input $G\delta[n]$. Now, note that such system is causal (uses only past information), and that the input is the impulse function. We know that if the system is causal, the impulse response of such system is causal too. Thus, $h[n]$ is a causal signal: $h[n]=0,\forall n&lt;0$.
Now, let's get into the autocorrelation function thing. From your procedure you have the following:
\begin{align}
\hat{R}(m)&amp;=\sum_{n=m}^{\infty}h[n-m]h[n]\\
\end{align}
So, you want to write this summation starting from $n=0$ right? If $m=0$ we are done. There are other two case: $m&lt;0$ and $0&lt;m$. In your procedure you did:
\begin{align}
\sum_{n=m}^{\infty}h[n-m]h[n] = \sum_{n=0}^{\infty}h[n-m]h[n] - \sum_{n=0}^{m-1}h[n-m]h[n]\\
\end{align}
However, this basically assumed that $0&lt; m$. Otherwise, the second summation does't make sense since the lower limit would be greater than the upper limit. What if $m&lt;0$? Then you would have:
$$
\sum_{n=m}^{\infty}h[n-m]h[n] = \sum_{n=0}^{\infty}h[n-m]h[n] + \sum_{n=m}^{-1}h[n-m]h[n]
$$
Lets inspect both cases under the causality property of $h[n]$. Hence, for the first case (with $0&lt; m$) we have:
$$
\sum_{n=0}^{m-1}h[n-m]h[n] = 0, \ \ \text{since } h[n-m]=0 \text{ for } n-m&lt;0
$$
this is true, since from the summation index range we have $0\leq n \leq m-1$ or equivalently $n&lt;m\implies n-m&lt;0$ (thus $h[n-m]=0$).
Now, for the second case (with $m&lt;0$) we have:
$$
\sum_{n=m}^{-1}h[n-m]h[n]=0 , \ \ \text{since } h[n]=0 \text{ for } n&lt;0
$$
this is true, since from the summation index range we have $m\leq n \leq -1$ or equivalently $n&lt;0$.
Thus, no matter what value of $m$ you have, the result is
$$
\hat{R}(m)=\sum_{n=0}^{\infty}h[n-m]h[n] = \hat{R}(-m)
$$
Note that this works due to the causality property of $h[n]$: this same reasoning works for any causal signal, regardless of its particular definition (as the one you gave with $h[n]$).