Find autocorrelation of exponential signal $a^nu[n]$
Let $x[n]=a^nu[n], |a|&lt;1$. Autocorrelation is 

$$\phi_{xx}[n]=\sum_{m=-\infty}^{\infty}x[m]x[m-n]=\sum_{m=-\infty}^{\infty}a^mu[m]a^{m-n}u[m-n]$$
First assume that $n&gt;0$. In this case, we have 

$$u[m]u[m-n]=\begin{cases}0,&amp; \forall m&lt;n\\
1,&amp; \forall m\ge n\end{cases}$$
Therefore,
$$\begin{align}
\phi_{xx}[n]&amp;=\sum_{\color{red}{m=n}}^{\infty}x[m]x[m-n]\\
&amp;=\sum_{m=n}^{\infty}a^ma^{m-n}\\
&amp;=a^n(1 + a^2 + a^4 +\cdots )\\
&amp;=\frac{a^n}{1-a^2}\end{align}$$



For $n&lt;0$:

$$u[m]u[m-n]=\begin{cases}0,&amp; \forall m&lt;0\\
1,&amp; \forall m\ge 0\end{cases}$$

$$\begin{align}
\phi_{xx}[n]&amp;=\sum_{\color{red}{m=0}}^{\infty}x[m]x[m-n]\\
&amp;=\sum_{m=0}^{\infty}a^ma^{m-n}\\
&amp;=a^{-n}(1 + a^2 + a^4 +\cdots )\\
&amp;=\frac{a^{-n}}{1-a^2}\end{align}$$

and since $\phi_{xx}[n]=\phi_{xx}[-n]$, we can write it for all $n$ as
$$\phi_{xx}[n]=\frac{a^{|n|}}{1-a^2},\ |a|&lt;1$$