Proof that $1 + \frac12 + \frac1{2^2} + ... + \frac1{2^{n-1}}$ converges to $2$ So I need to prove
$$x_n = 1 + \frac12 + \frac1{2^2} + ... + \frac1{2^{n-1}}, x_n \rightarrow 2$$
Here's my current work on it
$$x_n - \frac12x_n = \left(1 - \frac12\right) + \left(\frac12 - \frac14 \right) + ... + \left(\frac1{2^{n-1}} - \frac1{2^n}\right)$$
You can then cancel every term in the above sequence except for the first and last
$$x_n - \frac12x_n = 1 - \frac1{2^n}$$
$$\frac12x_n = 1 - \frac1{2^n}$$
$$x_n = 2 - \frac2{2^n}$$
The thing is now I'm not quite sure how to conclude the proof. You can say that $x_n$ is bounded above by $2$, but is that enough? Do I somehow need to prove that $2$ is also the least upper bound?
Any help would be appreciated.
 A: It's a geometric series, with first term $1$, and common ratio $\frac{1}{2}$.
A well-known (high-school) formula is for the sum of the first $n$-terms:
$$S_n = a\left(\frac{1-r^n}{1-r}\right)$$
If $a=1$ and $r=\frac{1}{2}$ then
$$S_n \ = \ 1\left(\frac{1-(1/2)^n}{1-1/2}\right) \ = \ 2-\frac{1}{2^{n-1}}$$
To show that $S_n \to 2$ as $n \to \infty$, you need to show that for all $\varepsilon > 0$, there exists an $N$ for which, if $n>N$ then $\left|S_n-2\right|<\varepsilon$.
\begin{eqnarray*}
\left|S_n - 2\right| &=& \left|\left(2-\frac{1}{2^{n-1}}\right) - 2\right| \\ \\
&=& \left|-\frac{1}{2^{n-1}}\right| \\ \\
&=& \frac{1}{2^{n-1}}
\end{eqnarray*}
Hence, $\displaystyle{|S_n-2| < \varepsilon \iff \frac{1}{2^{n-1}}< \varepsilon \iff n>1-\frac{\log \varepsilon}{\log 2}}$
A: Given an arbitrarily small $\epsilon >0$, find $n$ such that $2 - \left( 1 - {2 \over 2^n} \right) < \epsilon$.
With simple re-arrangement we find:  
$n > - \log_2 \epsilon - 1$.
In short, for an arbitrarily small $\epsilon$ we can always find an $n$ such that the series is closer to $2$ than $\epsilon$.  Moreover, it is trivial to prove that the series cannot converge to a number greater than 2.0.
