Answer
Yes.
$\frac{2^{n+1}-1}{2^{n}}$
Work Step by Step
Yes.
Simplify the right hand side as:
$RHS=\frac{1-\frac{1}{2^{n+1}}}{1-\frac{1}{2}}=\frac{2^{n+1}-1}{2^{n+1}-2^{n}}=\frac{2^{n+1}-1}{2^{n}}$
As $n$ is a non-negative integer, we have $2^{n+1}-1$ and $2^n\ne0$ as integers, thus $RHS$ is a rational number.
