Why does the comparison test fail here? Someone recently stumped me here.
Consider the series $\sum_3^{\infty} \dfrac{1}{x^{1.0001}}$ and the series $\sum_3^{\infty} \dfrac{1}{x\ln{x}}$.
The first series in convergent, and the second series is divergent. Yet, for all $x \geq 3$, $\dfrac{1}{x^{1.0001}} \geq \dfrac{1}{x\ln{x}}$. How can this be?
My conclusion is that the proposed inequality is actually false and so the direct comparison test cannot be applied, but the $x$ must be very large, I haven't been able to convince myself of this.
 A: Note that$$\frac1{x^{1.0001}}\geqslant\frac1{x\ln x}\iff x^{1.0001}\leqslant x\ln x\iff x^{0.0001}\leqslant\ln x,$$which is false if $x$ is large enough:$$\alpha>0\implies\lim_{x\to\infty}\frac{x^\alpha}{\ln x}=\infty.$$
A: It's a standard fact about limits that any positive power of $x$ (like $x^{0.0001}$ for instance) eventually outgrows $\ln x$, so $x^{1.0001} \gg x \ln x$ for all sufficiently large $x$.
A: So, the last inequality is wrong. And as you suspected, in fact $\frac{1}{x^{1.0001}}\leq \frac{1}{x\ln x}$ for $x$ very large. Indeed, this inequality is equivalent to $\ln x\leq x^{0.0001}$. Now, you may not be convinced yet because you may think $0.0001$ is very small. But if you choose an extremely large $x$, e.g. $x=e^{10^8}$, you have that $\ln x=10^8=100000000$ (sure is big), but $x^{0.0001}=e^{10^4}\geq 2^{10000}=16^{2500}\geq 10^{2500}=10000\dots 000$ (there are $2500$ zeros, crazy big!). I hope that convinces you that no matter how small is your number $\alpha>0$, you have that $\ln x\leq x^{\alpha}$ for $x$ big enough. 
