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This is from the UPenn prelim questions.

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 $x=0, (1-x)^2 f(x) = 0$, and $x=0$, and see how the value of the derivative of $(1-x)^2 f(x)$ 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 $0 < x < 1$, we're done, but the expression got way too messy while I was doing it, and I was really getting nowhere with it (in retrospect, it doesn't even look like a right method to do it).

Could someone help me with this? Any hints would be greatly appreciated.

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I'm done editing my answer I think. Hard to keep things neat and correct on a phone for me sometimes :) – snulty Aug 26 '14 at 7:31
up vote 1 down vote accepted

$f (x)=\sum \limits_{n=1}^\infty \frac {nx^n}{1-x^n} $

$(1-x)^2f(x)= \sum \limits_{n=1}^\infty \frac {n\cdot (1-x)\cdot x^n}{1+x+x^2+\ldots +x^{n-1} } $

Now consider each term recalling $x$ is non negative and the bottom line is maximal at one locally.

$ \frac {n\cdot (1-x)\cdot x^n}{1+x+x^2+\ldots +x^{n-1}} \geq \frac{n\cdot (1-x)\cdot x^n}{n} =x^n-x^{n+1}$.

Considering the series again we see it's telescoping so $(1-x)^2 f (x) \geq x$.

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$(1-x^2)f(x) \ge x$ means $f(x) - \frac{x}{1-x^2} \ge 0$

Now, $f(x) - \frac{x}{1-x^2} = \frac{x}{1-x} + \frac{2x^2}{1-x^2} - \frac{x}{1-x^2} + \cdot\cdot\cdot \ge \frac{x+x^2 + 2x^2 - x}{1-x^2} = 3\frac{x^2}{1-x^2} \ge 0$

I hope I didn't miss anything.

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thanks for the reply, but the question is about showing $(1-x)^2f(x) \geq x$, not $(1-x^2)f(x) \geq x$. – user98235 Mar 12 '14 at 7:36
Sorry, misread the question. Please disregard. – user58697 Mar 12 '14 at 7:53

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