This is from the UPenn prelim questions. http://hans.math.upenn.edu/amcs/AMCS/prelims/prelim_review.pdf
(I asked the question before, and there was no answer, so I am asking it again.. I'm not sure if this is allowed or not in this forum, and if this violates the rules, I will delete the post).
We have the following series
$$f(x) = \sum_{n=1}^{\infty} \frac{nx^n}{1-x^n}$$
It's easy to show that the series converges for all x in the interval (-1, 1) The hard part (At least for me) was to show that for x in [0, 1), we have the following inequality:
$$(1-x)^2 f(x) \geq x$$
I tried the following: when $x=0,(1−x)2f(x)=0$, and $x=0$, and see how the value of the derivative of $(1−x)2f(x)$ compares with 1 (which is the derivative of x), for $0<x<1$, then if that derivative is greater than or equal to 1 for 0
Could someone help me with this? Any hints would be greatly appreciated.
sequences-and-series
$\large f(x)=\sum_{n=1}^{\infty}\frac{nx^n}{(1-x^n)}$
Applying a Taylor expansion, we have
$\large f(x)=\sum_{n=1}^{\infty}nx^n(1+\sum_{k=1}^{\infty}x^{nk})\geq\sum_{n=1}^{\infty}nx^n$
as $0\leq x < 1$
The sum
$\large \sum_{n=1}^{\infty}nx^n=x\sum_{n=0}^{\infty}nx^{n-1}=x\frac{d}{dx}(\sum_{n=0}^{\infty}x^n)=x\frac{d}{dx}(\frac{1}{1-x})=\frac{x}{(1-x)^2}$
where we note that $\sum_{n=0}^{\infty}x^n$ is a sum to infinity of a Geometric progression, so that it equals $\frac{1}{1-x}$.
So we have
$\large f(x)\geq \frac{x}{(1-x)^2}$
Multiplying both sides by $(1-x)^2$, which is a positive value, thus keeping the same direction of the inequality, we have
$\large (1-x)^2f(x)\geq x$