Limit of $\sum_{r=1}^{n} \frac{r}{2^r}$ as $n\to \infty$ 
Consider the sequence $u_{n}=\sum\limits_{r=1}^{n} \frac{r}{2^r}$ with $n \ge 1$.
  Then the limit of $u_{n}$ as $ n \rightarrow \infty$ is ?

I actually treated each term as an AGP and got $$u_{n}=2-\frac{1}{2^{r+1}}-\frac{n}{2^{r+1}}$$ 
But how to get the limit?

This is a well-known result due to Nicole Oresme
\begin{align}
\sum_{n=1}^\infty \frac{n}{2^n}=2
\end{align}
Here is the proof 
Consider the function
\begin{align*}
f(z)=\frac{1}{1-z}
\end{align*}
The power series expansion of $f(z)$ is
\begin{align}
f(z)=\sum_{n=1}^{\infty}z^n
\end{align}
Now take the derivative of $f$, which is
\begin{align*}
f'(z)=\frac{1}{(1-z)^2}=\sum_{n=1}^{\infty}nz^{n-1}
\end{align*}
Multiplying by $z$
\begin{align}
zf'(z)=\frac{z}{(1-z)^2}=\sum_{n=1}^{\infty}nz^{n}\tag{1}
\end{align}
Substituting $z=1/2$ in (1)
\begin{align}
\sum_{n=1}^\infty \frac{n}{2^n}=\frac{\frac{1}{2}}{\biggl(1-\frac{1}{2}\biggl)^2}=2
\end{align} 
