# 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 $0 < x < 1$

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

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...

• My hint: differentiate term-by-term and see what happens. ... in general, knowing closed forms for $\sum a_n$ and for $\sum b_n$ does not give you a closed form for $\sum a_nb_n$. Apr 7, 2017 at 10:59
• you should look into generating functions. A method for solution is differentiating to solve for a diff equation. They are incredibly powerful and fun Apr 7, 2017 at 11:12
• @GEdgar, Wow, I never thought that this will be a way! okay, So, i differentiated and got this, $\frac{d^2S}{ dx^2} = \frac{1}{1-x}$, Where $S$ is denotes the sum of infinite series. Should I integrate it to get $S$ ? Apr 7, 2017 at 11:16

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 $$r=(x/(n+1))/(1/(n-1))$$