Verify proof that $x_n = \sum_{k=1}^n {k \over 2^k}$ is bounded and find its supremum and infinum The problem I'm solving states:

Let $n\in \mathbb N$ and $x_n$ be a sequence:
  $$
x_n = \sum_{k=1}^n {k \over 2^k}
$$ 
  Prove $x_n$ is bounded and find $\sup\{x_n\}$ and $\inf\{x_n\}$

Let $S_n$ denote the sum, then:
$$
S_n = \frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +  \dots + \frac{n}{2^n}
$$
Multiply $S_n$ by $1 \over 2$:
$$
{1\over 2} S_n = \frac{1}{2^2} + \frac{2}{2^3} + \frac{3}{2^4} +  \dots + \frac{n}{2^{n+1}}
$$
Subtract the sums:
$$
\begin{align}
S_n - { 1 \over 2 } S_n &= \sum_{k=1}^n {1\over 2^k} - {n\over 2^{n+1}} = \\
&= 1 - {1 \over 2^n } - {n \over 2^{n+1}} = \\
&= 2 - {n \over 2^n} - {2 \over 2^n} < 2
\end{align}
$$
Therefore:
$$
\inf\{x_n\} = {1\over 2} \le x_n < 2 = \sup\{x_n\} \\
$$
Have I done it the right way?

It's may be easier to do it this way: since $(x_n)_n$ is strictly increasing, $\inf x_n=x_1=1/2$ and $\sup x_n=\lim x_n$, which you can calculate by noting that $f(x)=\sum_{n=0}x^n=\frac{1}{1-x};\ |x|<1.$ 
Now, $f$ may be differentiated inside its interval of convergence, so $x\sum_{n=1}nx^{n-1}=\frac{x}{(1-x)^{2}}.$ To finish, evaluate this expression at $x=1/2.$
