About the infinite geometrical sequence factored with n

Question

I just came across this thread, and i asked myself:

I know that $\sum^\infty_{n=0} x^n = \frac{1}{1-x}$

But what happens when we set up the sum like

$$\sum^\infty_{n=0} nx^n = ?$$

There is probably a formula for that as well, at least there must be some coincidence, for the limited sums i already saw the prove in the thread i mentioned above. My books could not help me out so far.. If there is a formula for this, i would also like to see the derivation of it..

Searched the web with no luck so far. If you know a website or thread where this is explained, probably a link is enough for an answer.. I appreciate your help!

Answer 1

Note that for $|x| < 1$, we have

$$\sum_{n=0}^\infty nx^n = x \sum_{n=0}^\infty nx^{n-1} = x \sum_{n=0}^\infty \frac{d\ }{dx} x^n = x \frac{d\ }{dx} \sum_{n=0}^\infty x^n = x \frac{d\ }{dx} \frac{1}{1-x} = \frac{x}{(1-x)^2}$$

You can now write down an expression for your sum.

Answer 2

Differentiate the first expression with respect to $x$, then multiply by $x$, then add missing initial term(s)?

Answer 3

I don't know if this is quite what you're looking for, but I know I have used this one is class before and if'd you'd like to look I've also used it in a question that i posted on here. Hope it helps!

$$\sum_{k=1}^nku^k={u\over (1-u)^2}\bigg[nu^{n+1}-(n+1)u^n+1\bigg]\forall u\ge 1$$

P.S.- I know it's not for $k=0$ to $n$, but i thought it might help still.