Summations

$\begin{aligned} \sum_{N=1}^\infty x^N & = \frac x{1-x} & \small \color{#3D99F6}{\text{for }x<1} \\ \frac d{dx} \sum_{N=1}^\infty x^N & = \frac d{dx}\left(\frac x{1-x}\right) & \small \color{#3D99F6}{\text{Differentiate both sides w.r.t }x} \\ \sum_{N=1}^\infty Nx^{N-1} & = \frac 1{(1-x)^2} & \small \color{#3D99F6}{\text{Multiply both sides with }x} \\ \sum_{N=1}^\infty Nx^{N} & = \frac x{(1-x)^2} & \small \color{#3D99F6}{\text{Putting }x=\frac 1X} \\ \sum_{N=1}^\infty \frac N{X^N} & = \frac {\frac 1X}{\left(1-\frac 1X\right)^2} \\ \implies \frac x{(1-x)^2} & = \frac {10}9 \\ 9x & = 10(1-x)^2 \\ 10x^2 - 29x + 10 & = 0 \\ (5x-2)(2x-5) & = 0 \\ \implies x & = \frac 25 & \small \color{#3D99F6}{\text{Note that }x<1} \\ X & = \frac 52 = \boxed{2.5} \end{aligned}$

$Let$ $X > 1.$ $$ $\sum_{N = 1}^{\infty} \frac{N}{X^N}$ = $\frac{1}{X} *$ $(1 + \frac{1}{X}$ + $\frac{1}{X^2}$ $+ ...) +$ $\frac{1}{X^2} *$ $(1 + \frac{1}{X}$ + $\frac{1}{X^2}$ $+ ..) + ... =$ $\frac{X}{X - 1}$ * $\frac{1}{X} *$ $\frac{X}{X - 1} =$ $\frac{X}{(X - 1)^2} = \frac{10}{9}$ $\implies$ $10x^2 - 29x + 10 = 0$ $\implies$ $X = 5/2,2/5.$ $$ $For$ $X > 1$ $the$ $positive$ $series$ $\sum_{N = 1}^{\infty} \frac{N}{X^N}$ $converges$ $$ $$ $\therefore$ $choose$ $X = 5/2.$

$\sum_{N=1}^\infty \dfrac N{X^N} = \frac{1}{x} + \frac{2}{x^2} + \frac{3}{x^3} + ... = \frac{10}{9}$

$\frac{1}{x} + (\frac{1}{x^2} + \frac{1}{x^2}) + (\frac{1}{x^3} + \frac{1}{x^3} + \frac{1}{x^3}) + ... = \frac{10}{9}$

$( \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + ...) + ( \frac{1}{x^2} + \frac{1}{x^3} + \frac{1}{x^4} + ...) + ( \frac{1}{x^3} + \frac{1}{x^4} + \frac{1}{x^5} + ...) + .... = \frac{10}{9}$

$\frac{\frac{1}{x}}{1-\frac{1}{x}} + \frac{\frac{1}{x^2}}{1-\frac{1}{x}} + \frac{\frac{1}{x^3}}{1-\frac{1}{x}} + ... = \frac{10}{9}$

$\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + ... = \frac{10}{9} (1 - \frac{1}{x}) \boxed{EPMD}$

$\frac{\frac{1}{x}}{1 - \frac{1}{x}} = \frac{10}{9} (1 - \frac{1}{x})$

After a little simplification, the equation becomes a quadratic equation with roots $\frac{2}{5}, \frac{5}{2}$ . We take $\boxed{x=\frac{5}{2}}$ since $x>1$ is specified in the question.