<s>[INST] Summation to n terms of series \begin{gather}
In\ a\ problem\ given\ to\ us\ by\ our\ teacher,\ one\ part\ of\ it\ requires\ the\ summation\ of\ the\  \notag\\
following\ sequence\ to\ a\ finite\ number\ of\ terms\ ( n) , \notag\\
t_{r\ } \ =\ \frac{1}{r\times 2^{r}} \notag\\
 \notag\\
S\ =\ \frac{1}{1\times 2} +\frac{1}{2\times 2^{2}} +\frac{1}{3\times 2^{3}} +....\ \frac{1}{r\times 2^{r}} \notag\\
 \notag\\
I\ tried\ the\ following\ approach\ :\  \notag\\
 \notag\\
\ S\ =\ \frac{1}{1\times 2} +\frac{1}{2\times 2^{2}} +\frac{1}{3\times 2^{3}} +....\ \frac{1}{r\times 2^{r}} \ \ \ ( i) \notag\\
\ \frac{S}{2} \ =\ \frac{1}{1\times 2^{2}} +\frac{1}{2\times 2^{3}} +\frac{1}{3\times 2^{4}} +....\ \frac{1}{( r-1) \times 2^{r}} \ +\ \frac{1}{r\times 2^{r+1}} \ \ \ \ ( ii) \notag\\
( i) \ -\ ( ii) \  \notag\\
\frac{S}{2} \ =\ \frac{1}{1\times 2} \ -\left( \ \left( 1-\frac{1}{2}\right) 2^{-1} \ +\ \left(\frac{1}{2} \ -\ \frac{1}{3}\right) 2^{-2} \ +...\left(\frac{1}{r-1} -\frac{1}{r}\right) 2^{-r}\right) \ -\ \frac{2^{-r}}{r}\\
 \notag\\
\frac{S}{2} \ =\ \frac{1}{1\times 2} \ -\left( \ \left(\frac{1}{2}\right) 2^{-1} \ +\ \left(\frac{1}{6}\right) 2^{-2} \ +...\left(\frac{1}{r( r-1)}\right) 2^{-r}\right) \ -\ \frac{2^{-r}}{r} \notag\\
Here\ I\ got\ stuck\ ,\ as\ the\ series\ generated\ above\ again\ generates\ another\ similar\ sequence.\  \notag\\
 \notag\\
 \notag\\
 \notag
\end{gather}
I don't think the sum telescopes as well. Can calculus be used in some way to calculate the sum?
I'll just add the actual problem, in case it offers some help :
\begin{gather*}
t_{r} \ =\ \frac{r+2}{r( r+1)} \ 2^{-r}\\
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\end{gather*}
 [/INST] Considering the actual probem 
$$t_{r}  =\frac{r+2}{r( r+1)} x^r=2\frac{x^r}{r}-\frac{x^r}{r+1}=2\frac{x^r}{r}-\frac 1x\frac{x^{r+1}}{r+1}$$
$$\sum_{r=1}^n t_r=2\sum_{r=1}^n \frac{x^r}{r}-\frac 1x\sum_{r=1}^n\frac{x^{r+1}}{r+1}$$ Differentiate both sides with respect to $x$
$$\left(\sum_{r=1}^n t_r \right)'=2\sum_{r=1}^n x^{r-1}-\frac 1x\sum_{r=1}^n{x^{r}}$$
I am sure that you can take it from here.
At the end, make $x=\frac 12$ and $n\to \infty$
</s>