<s>[INST] Find the sum of this series May I know how I should go about finding the sum of this series? 
$\displaystyle\sum_{n=1}^\infty$ $\dfrac{n}{2^{n-1}}$
I am really stuck. Thanks!
 [/INST] Here is a way without differentiation.
Let
$$S=\sum_{n=1}^\infty \frac{n}{2^{n-1}}=\frac{1}{1}+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\dotsb$$
Now, let's see if we can get $\displaystyle\sum_{n=1}^\infty \frac{n-1}{2^{n-1}}$ in terms of $S$ in two different ways.
First way:
\begin{align}
\sum_{n=1}^\infty \frac{n-1}{2^{n-1}}&=\sum_{n=1}^\infty \frac{n}{2^{n-1}}-\sum_{n=1}^\infty \frac{1}{2^{n-1}} \\
&=S-\left(1+\frac{1}{2}+\frac{1}{4}+\dotsb\right) \\
&=S-\frac{1}{1-\frac{1}{2}}=S-2
\end{align}
Second way:
\begin{align}
\sum_{n=1}^\infty \frac{n-1}{2^{n-1}}&=\frac{1}{2}\sum_{n=1}^\infty \frac{n-1}{2^{n-2}} \\
&=\frac{1}{2}\left(0+\frac{1}{1}+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\dotsb\right) \\
&=\frac{1}{2}S
\end{align}
Since these must be equal, we have:
$$S-2=\frac{1}{2}S$$
Solving, we get $S=4$, which is the answer.
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