Summation of fraction ^ n * n

I don't know of any name, but here's a derivation:

So I’m starting to see that summation n = 0 to infinity of (a/b)^n *n = ab/(b-a)^2 , (where b>a so it actually converges )...? This is actually true for all cases, and a known result? If so, I’d like to know what you’d actually call it so I can look up the proof as it’s never been mentioned or taught at my Uni, although there’s been worked solutions using this... Thanks

\(\text{Let $x = \dfrac{a}{b} \quad b > a\quad $ or $\quad|x|<1$ }\)

Now we have:

\(\displaystyle \sum \limits_{n=0}^{\infty} nx^n\)

Let the progression to be summed be put equal to s:

\(s = 0+x+2x^2+3x^3+4x^4+\cdots + nx^n +\cdots\)

It is divided by x and multiplied by dx then

\(\dfrac{s\ dx}{x} = dx +2x\ dx+3x^2\ dx+4x^3\ dx+\cdots + nx^{n-1}\ dx +\cdots \)

and with the integrals taken this equation is found

\(\displaystyle \int \dfrac{s\ dx}{x} = x +x^2+x^3+x^4 +\cdots + x^n + \cdots = \dfrac{x}{1-x} \quad \text{ infinite geometric progression }\)

Therefore from the equation:

\(\displaystyle \int \dfrac{s\ dx}{x} = \dfrac{x}{1-x}\)

on differentiation s is found. For the equation becomes:

\(\displaystyle \dfrac{s}{x} = \dfrac{1}{1-x} + x(-1)(1-x)^{-2}(-1) \\ \displaystyle \dfrac{s}{x} = \dfrac{1}{1-x} + \dfrac{x}{(1-x)^2} \\ \displaystyle \dfrac{s}{x} = \dfrac{1-x+x}{(1-x)^2} \\ \displaystyle \dfrac{s}{x} = \dfrac{1 }{(1-x)^2} \)

thus there is produced:

\(\displaystyle s = \dfrac{x}{(1-x)^2} \)

So \(\text{let $x = \dfrac{a}{b} $}:\)

\(\begin{array}{|rcll|} \hline s &=& \dfrac{ \dfrac{a}{b} } {\left(1-\dfrac{a}{b} \right)} \\\\ &=& \dfrac{ a } {b \dfrac{(b-a)^2}{b^2} } \\\\ \mathbf{s} & \mathbf{=} & \mathbf{\dfrac{ ab } { (b-a)^2 }} \\ \hline \end{array}\)

Finally

\(\displaystyle \sum \limits_{n=0}^{\infty} n \left(\dfrac{a}{b}\right)^n = \dfrac{ ab } { (b-a)^2 } \)