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...

  • 3
    $\begingroup$ 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 $\sum a_nb_n$. $\endgroup$
    – GEdgar
    Apr 7, 2017 at 10:59
  • $\begingroup$ you should look into generating functions. A method for solution is differentiating to solve for a diff equation. They are incredibly powerful and fun $\endgroup$ Apr 7, 2017 at 11:12
  • 1
    $\begingroup$ @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 $S$ ? $\endgroup$ Apr 7, 2017 at 11:16

2 Answers 2


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

  • $\begingroup$ No – you can't. $\endgroup$ Apr 7, 2017 at 12:29
  • $\begingroup$ Oh ok. I guess i messed up somewhere. Sorry :| $\endgroup$
    – Shash
    Apr 7, 2017 at 12:38

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