$$a_n = \frac{1}{2^n}, \qquad \sum_{n=1}^{\infty} a_n = \lim_{n\to\infty} (1-\frac{1}{2^n}) = 1$$
Depends how you interpret it. First let us look at the solution to
First we will be using (for $r<1$)
$$
\sum_{k=0}^{n} ar^{k} = \frac{a- ar^{n+1}}{1-r}
$$
$$
\sum_{k=0}^\infty ar^k = \frac{a}{1-r}
$$
Now in the problem above
$$a_n = \frac{1}{2^n}$$ and therefore
$$
\begin{align*}
\sum_{n=1}^{\infty} a_n &= \sum_{n=1}^{\infty}\frac{1}{2^n}\\
&= {\frac{1}{2}} ( 1 + \frac{1}{2^{1}} + \frac{1}{2^{2}} + \dots )\\
&= {\frac{1}{2}} (\frac{1}{1-\frac{1}{2}})
&= 1
\end{align*}
$$
And
$$\lim_{n\to\infty} (1-\frac{1}{2^n}) = 1 - \lim_{n\to\infty} \frac{1}{2^n} = 1$$