Find the value of this series

Consider the Maclaurin series of an infinite geometric progression.

$\begin{aligned} \sum_{n=0}^\infty x^n & = \frac{1}{1-x} \quad \quad \small \color{#3D99F6}{\text{Differentiate both sides with respect to }x} \\ \sum_{n=1}^\infty n x^{n-1} & = \frac{1}{(1-x)^2} \quad \quad \small \color{#3D99F6}{\text{Multiply both sides by }x} \\ \sum_{n=1}^\infty n x^n & = \frac{x}{(1-x)^2} \quad \quad \small \color{#3D99F6}{\text{Put }x=\frac{1}{2}} \\ \sum_{n=1}^\infty \frac{n}{2^n} & = \frac{\frac{1}{2}}{\left(1-\frac{1}{2}\right)^2} = \boxed{2} \end{aligned}$

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

Using sum of infinite terms of AGP.

$\Rightarrow \dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3}{8}+...$

$\dfrac{1}{2}\left(1+\dfrac{1}{2}+\dfrac{2}{4}+...\right)$

Now, $a=1$ , $r=\dfrac{1}{2}$ and $d=1$ .

$\dfrac{1}{2}\left(\dfrac{1}{1-\frac{1}{2}}+\dfrac{1×\frac{1}{2}}{\left(1-\frac{1}{2}\right)^2}\right)$

$\dfrac{1}{2}\left(2+\dfrac{1}{2}×4\right)$

$\dfrac{1}{2}×4=\boxed{2}$