Condition for $\sum_{x=1}^{\infty} f(x)\left(\frac{1}{f(x)-1}\right)^{2x+1} < \infty\,?$

Question

Under what conditions on $f(x)$ is $$\sum_{x=1}^{\infty} f(x)\left(\frac{1}{f(x)-1}\right)^{2x+1} < \infty\,?$$

We can assume that $f(x)$ is a smooth increasing function of $x$.

It is clearly true when $f(x) \geq x$.

Answer 1

I will show it converges iff $\forall x f(x)\leq 0$ or $\forall x\geq x_0 :f(x) > 2$.

If $\forall x f(x)\leq 0$ then this clearly holds (might go to $-\infty$).

If $\forall x\geq x_0 :f(x) > 2$:

$$\sum_{x=1}^{\infty} f(x)\left(\frac{1}{f(x)-1}\right)^{2x+1} = C+\sum_{x=x_0}^{\infty} \frac{f(x)}{f(x)-1}\left(\frac{1}{f(x)-1}\right)^{2x} \leq C+ \sum_{x=x_0}^{\infty} 2\left(\frac{1}{f(x)-1}\right)^{2x} \leq C+ \sum_{x=x_0}^{\infty} 2\left(\frac{1}{c}\right)^{2x}$$, which clearly converges (as $c$>1).

If this doesn't hold, then $$\sum_{x=1}^{\infty} f(x)\left(\frac{1}{f(x)-1}\right)^{2x+1} \geq \sum_{x=1}^{\infty} \frac{f(x)}{f(x)-1}\left(\frac{1}{f(x)-1}\right)^{2x} \geq \sum_{x=1}^{\infty} \left(\frac{1}{f(x)-1}\right)^{2x}\ge \sum_{x=1}^{\infty} \left(\frac{1}{\bar c}\right)^{2x}= \infty$$ , since $\bar c\leq 1$