$\displaystyle \sum\limits_{n = 0}^{n - 1} \left({\frac {1}{2}}\right) ^ n$
I assume that $\displaystyle n-1$ on sigma is a constant and not related to the summation.
$\displaystyle \sum _{k=0}^a r^k = \frac{r^{1+a}-1}{r-1}$
So,
$\displaystyle \sum_{n = 0}^{n - 1} \left({\frac {1}{2}}\right)^n = \frac{\left(\frac{1}{2}\right)^{1+(n-1)}-1}{\frac{1}{2}-1} = 2 - 2^{1-n}$
If it was the infinite series $\displaystyle \sum_{n=1}^{\infty} \left( \frac{1}{2} \right)^n$ then the answer would indeed be 1.
But it's not so it's not.
(Note: The very first term of the given series is $\displaystyle \left( \frac{1}{2} \right)^0 = 1$ so clearly it can't sum to 1 .......)