Let $x[n]=a^nu[n], |a|<1$. Autocorrelation is

$$R_{xx}[m]=\sum_{n=-\infty}^{\infty}x[n]x[n-m]=\sum_{n=-\infty}^{\infty}a^nu[n]a^{n-m}u[n-m]$$ First assume that $m>0$. In this case, we have

$$u[n]u[n-m]=\begin{cases}0,& \forall n<m\\ 1,& \forall n\ge m\end{cases}$$ Therefore, $$\begin{align} R_{xx}[m]&=\sum_{\color{red}{n=m}}^{\infty}x[n]x[n-m]\\ &=\sum_{n=m}^{\infty}a^na^{n-m}\\ &=a^m(1 + a^2 + a^4 +\cdots )\\ &=\frac{a^m}{1-a^2}\end{align}$$ and since $R_{xx}[m]=R_{xx}[-m]$, we can write it for all $m$ as $$R_{xx}[m]=\frac{a^{|m|}}{1-a^2},\ |a|<1$$

Let $x[n]=a^nu[n], |a|<1$. 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 $n>0$. In this case, we have

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

For $n<0$:

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

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

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