Calc limit of a series My HW is
$$\lim\limits_{n\to\infty}\dfrac{1\cdot x+2\cdot x^2+3\cdot x^3+\dots+n\cdot x^n}{n\cdot x^{n+1}}$$ as $$x\gt1$$
I've multiplied both numerator and denominator by $$1/x^n$$
so my answer $$1/x$$ 
I was told that it is not correct.
Can anybody please help? 
 A: $$\lim\limits_{n\to\infty}\dfrac{1\cdot x+2\cdot x^2+3\cdot x^3+\dots+n\cdot x^n}{n\cdot x^{n+1}}$$ as $$x\gt1$$
Here it is $\frac{\infty}{\infty}$, let us use L'Hôpital's rule.
If I differentiate $n$-times (the least surviving order) and then take $n->\infty $, will solve it.
$$\lim\limits_{n\to\infty}\Big(\dfrac{n\cdot n\cdot(n-1)\cdot(n-2)\dots1}{n\cdot(n+1)\cdot n\cdot(n-1)\dots 1\cdot x^{1}} = \dfrac{1}{(n+1)x} \Big) \rightarrow 0 $$
A: Rewrite the numerator as
$$
\sum_{i=1}^n \sum_{j=0}^{n-i} x^j = \sum_{i=1}^n \frac{x^{n-i}-1}{x-1}=\frac{1}{x-1}\left(\frac{x^n-1}{x-1}-\frac{n(n+1)}{2}\right).
$$
It follows that the limit is equal to
$$
\lim_{n\to \infty}\frac{1}{nx^{n+1}}\left(\frac{x^n-1}{(x-1)^2}-\frac{n(n+1)}{2(x-1)}\right).
$$
Both summands tend to $0$.
A: Remember the geometric sum?
$$1+x+x^2+\dots+x^n=\frac{1-x^{n+1}}{1-x}$$
Differentiate both sides and you'll get
$$1+2x+3x^2+4x^3+\dots+nx^{n-1}=\frac d{dx}\frac{1-x^{n+1}}{1-x}$$
Then multiply both sides by $x$.
$$1x+2x^2+3x^3+\dots+nx^n=x\left(\frac d{dx}\frac{1-x^{n+1}}{1-x}\right)$$
then the rest is some algebra and cancellation.
