Finding sum $\frac{1}{2} x^2 + \frac{1}{2} \frac{1}{3} x^3 + \frac{1}{4} \frac{1}{3} x^4 + ... $ I need to find following: for $0 &lt; x &lt; 1$

$$\frac{1}{2} x^2 + \frac{1}{2} \frac{1}{3} x^3 + \frac{1}{4} \frac{1}{3} x^4  + ... $$

My attempt: 

I can see that the sum is composed of two infinite sums, one is $\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \sum_{n=1}^{\infty} \left( \frac{1}{n} -
 \frac{1}{n+1} \right)$ (Telescoping ) and another is, $\sum_{n=2}^{\infty} x^n$ (it's a G.P.). How can I use these for solving the sum in question?

Any hints will be appreciated...
Let us set
$$
S(x)=\frac{1}{2} x^2 + \frac{1}{2} \frac{1}{3} x^3 + \frac{1}{4} \frac{1}{3} x^4  + \ldots 
$$
then
$$
\frac{dS(x)}{dx}=x + \frac{1}{2}x^2 + \frac{1}{3} x^3  + \ldots. 
$$
This is a well-known Taylor series being
$$
   x + \frac{1}{2}x^2 + \frac{1}{3} x^3  + \ldots = -\ln(1-x).
$$
We now note that $S(0)=0$, then we have to solve the simple differential equation
$$
   \frac{dS(x)}{dx}=-\ln(1-x)
$$
with the given initial condition. This yields
$$
  S(x)=(1-x)\ln(1-x)+x
$$
that is the result you were looking for.
Hint
U can make the series into a geometric progression where $$a=\frac{x^2}{2}$$  and $$r=(x/(n+1))/(1/(n-1))$$