A possible exception to the direct comparison test for series? $$\sum_{n=2}^\infty \frac1{n\ln n}$$   diverges by the integral test and p-series. 
But, 
$$\sum_{n=2}^\infty \frac1{n^{1.1}}$$ converges by p-series, and is greater than the first series, right? So, by the direct comparison test for series, the first series should converge. 
I know that a divergent series can't be smaller than a convergent series, so this must be wrong somehow. I think that
$$\frac1{n\ln n}  <       \frac1{n^{1.1}}$$
must be wrong, but I don't know how to show it. From what I can tell, the first series is larger at first, but at a very large n, is much smaller. 
Thanks for the help!! 
 A: Sorry to burst your bubble, but you're wrong about
$$\frac{1}{n\ln(n)}\lt \frac{1}{n^{1.1}}$$
Notice that for positive integer $n$, this statement is equivalent to
$$\ln(n)\gt n^{0.1}$$
First of all, this is false for $n=1,2,3$, but that doesn't matter, since we're looking at asymptotic domination here. So we should instead consider whether or not
$$\ln(n)\gt n^{0.1}$$
is true for large $n$. If this is true for large $n$, then the value of the limit
$$\lim_{n\to\infty} \frac{\ln(n)}{n^{0.1}}$$
should be $\infty$. However, by L'Hopital,
$$\lim_{n\to\infty} \frac{\ln(n)}{n^{0.1}}=\lim_{n\to\infty} \frac{\frac{1}{n}}{0.1n^{-0.9}}$$
$$\lim_{n\to\infty} \frac{\ln(n)}{n^{0.1}}=\lim_{n\to\infty} \frac{10}{n^{0.1}}$$
$$\color{red}{\lim_{n\to\infty} \frac{\ln(n)}{n^{0.1}}=0}$$
Which disproves your claim that
$$\frac{1}{n\ln(n)}\lt \frac{1}{n^{1.1}}$$
A: Note the inasmuch as $\log(n)\le n-1<n$, and $\log(n^{0.1})=0.1\log(n)$, we see that
$$\sum_{n=2}^N \frac1{n\log(n)}\ge 0.1\sum_{n=2}^N \frac{1}{n^{1.1}}\tag 1$$
One cannot conclude from $(1)$ that the partial sums on the left-hand side converges (they do not) as a consequence of convergence of the right-hand side of $(1)$.
A: Perhaps it is more clear to consider $n=e^{x_n}$ so that we get
$$\frac1{\ln(n)}=\frac1{x_n}$$
$$\frac1{n^{0.1}}=\frac1{e^{0.1x_n}}$$
$$\frac1{x_n}\quad\text{vs.}\quad\frac1{e^{0.1x_n}}$$
Which one converges to zero faster? Hopefully you realize exponential growth dominates linear growth.
