Arithmatico-geometric series

The given expression is:

$1+2x+3x^2+4x^3+\dots = 4$

Multiplying both sides by $x$ ,

$x+2x^2+3x^3+\dots = 4x$

Subtracting the second equation from the first,

$1+x+x^2+x^3+\dots = 4-4x$

The left-hand side is a geometric series and so:

$\dfrac{1(1-x^{\infty})}{1-x}=4-4x$

Since $x \in (0,1)$ , $x^{\infty} = 0$ :

$\dfrac{1}{1-x}=4-4x$

On simplification,

$4x^2-8x+3=0$

Which yields the roots $x= \dfrac{1}{2}$ and $x=\dfrac{3}{2}$ .

But since $x \in (0,1)$ ,

$x=\boxed{\dfrac{1}{2}}$

$\sum^{\infty}_{n=1} nx^{n-1} = 1 + 2x + 3x^2 + 4x^3 + ... =$ $\sum^{\infty}_{n=1} x^n = x + x^2 + x^3 + x^4 + ...$ $\frac{d}{dx}\left(x + x^2 + x^3 + x^4 + ... \right) = 1 + 2x + 3x^2 + 4x^3 + ...$ $\Rightarrow \frac{d}{dx}\sum^{\infty}_{n=1}x^n = \sum^{\infty}_{n=1}nx^{n-1}$ $\sum^{\infty}_{n=1}x^n = \frac{x}{1-x} \Rightarrow \frac{d}{dx}\left(\frac{x}{1-x}\right) = \sum^{\infty}_{n=1}nx^{n-1}$ $\frac{d}{dx}\left(\frac{x}{1-x}\right) = \frac{1}{\left(1-x\right)^2}$ $\frac{1}{\left(1-x\right)^2} = \sum^{\infty}_{n=1}nx^{n-1} = 4$ $\frac{1}{4} = \left(1-x\right)^2 \Rightarrow \pm \frac{1}{2} = 1-x$ $x = 1 \pm \frac{1}{2}$ $x<1 \Rightarrow x = \left(1 - \frac{1}{2} \right) = \frac{1}{2} = 0.5$

I literally just said "well, it has to have a nice solution. So how about a half?" I love it when guessing pays off. XD

$S = 1+2x+3x^{2}+4x^{3}+5x^{4} +........$

$xS = 0+x+2x^{2}+3x^{3}+4x^{4} +........$

$(1-x)S = 1+ x+x^{2}+x^{3}+x^{4} + ..........$

Since $0<x<1$ , hence sum of GP on right is $1/(1-x)$

$a+ar^{2}+ar^{3}+...... = a/(1-r)$

So $(1-x)S = 1/(1-x)$

$S = 1/(1-x)^{2} = 4$ as given

which will give $x = 0.5$ as the satisfactory value