I am confused about starting index of unit step function Let $$u[n]=\begin{cases}
1&\text{if $n\geq0$}\\
0&\text{if $n<0$}
\end{cases}
$$
Then $$u[n-1]=\begin{cases}
1&\text{if $n\geq1$}\\
0&\text{if $n<1$}
\end{cases}.$$
If we have an equation that contains something like $$\displaystyle\frac{1}{2^{n}}u[n]-\frac{1}{3^{n}}u[n-1]$$
If we wish to get rid of the unit step function, we shall assume $n\geq1$. My question is, would this mean that $u[n]=0$? To clarify my point if we start at $n=1$ and upwards, does that mean $u[n]=0$ if $n=0$? since we are starting from $n=1$ and above

The unit sep, $u(n)$ is equal to $1$ wheneve its argument is non-negative.  So, for $n\ge 1$, $u(n)=1$.  Moreover, we have
$$\frac{1}{2^n}u(n)-\frac{1}{3^n}u(n-1)=\begin{cases}\frac{1}{2^n}-\frac{1}{3^n}&,n\ge1\\\\
\frac1{2^n}&,1>n\ge 0\\\\0&,0>n\end{cases}$$
