Converge of the sum $\sum_{k=1}^{n} k x^k $ For what values ​​of x the sum converges  and what is the limit when $n \rightarrow \infty$
$\sum_{k=1}^{n} k x^k $
My work: 
First i try to calculate the interval and radius of convergence of $\sum_{k=1}^{n} k x^k $ by using the ratio test:
$$p(x) =\lim_{k \to \infty} \left| \frac{(k+1) x^{k+1}}{kx^k} \right| = \lim_{k\rightarrow \infty}\left|\frac{\left( k+1 \right)}{k}x\right|$$
here i stuck can you help me please.
 A: You are almost there.
$$\lim_{k \rightarrow \infty} \left| \frac{(k+1)}{k} x \right|= \lim_{k \rightarrow \infty}\left|\frac{k+1}{k}\right|.|x| = |x|,$$
now, for convergence we need ...?
To find the limit, you should follow the hint given by T.Bongers. 
In a bit more detail, we know that for $|x|<1$ we have 
$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \dots$$
Now differentiate both sides with respect to $x$ (valid inside the circle of convergence):
$$\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 + \dots$$
so 
$$\frac{x}{(1-x)^2} = x + 2x^2 + 3x^3 + 4x^4 + \dots$$
A: For $\left| x \right|<1$ yu have the absolute convergent series
$$\sum_{k=0}^{\infty}x^k = \frac{1}{1-x}$$ 
Because of the absolute convergence you are allowd to interchange summation with differentiation and you get
$$\frac{d}{dx}\sum_{k=0}^{\infty}x^k = \sum_{k=0}^{\infty}\frac{d}{dx}x^k  = \sum_{k=0}^{\infty} k\cdot x^{k-1} $$
On the other hand you have
$$\frac{d}{dx}\sum_{k=0}^{\infty}x^k = \frac{d}{dx} \frac{1}{1-x} = \frac{1}{\left(x-1\right)^2}$$ 
Thus it holds
$$\sum_{k=0}^{\infty} k\cdot x^{k-1} = \frac{1}{\left(x-1\right)^2}$$ 
Multiplying with x gives you
$$\sum_{k=0}^{\infty} k\cdot x^{k} = \frac{x}{\left(x-1\right)^2}$$ 
Note that the first summand on the left side is zero for $k=0$ so you have finally
$$\sum_{k=1}^{\infty} k\cdot x^{k} = \frac{x}{\left(x-1\right)^2}$$ 
