Does $\sum^{\infty}_{n=1}\frac{\ln{n}}{n^{1.1}}$ converge or diverge? Does $$\sum^{\infty}_{n=1}\frac{\ln{n}}{n^{1.1}}$$ converge or diverge? I think the basic comparison works but I have a hard time finding a comparer. Could someone suggest one?  
 A: The important thing here is that $\ln n \ll n^\alpha$ for any $\alpha > 0$, for $n$ large enough. You can check this by verifying the limit
$$ \lim_{n \to \infty} \frac{\ln n}{n^\alpha} = 0.$$
Let's see how we might use this to show that
$$ \sum_{n \geq 1} \frac{\ln n}{n^{1.1}}$$
converges. From above, there is some $N$ such that for all $n > N$, we have that $\ln n < n^{0.05}$, say. Then you can break your sum into two pieces,
$$ \sum_{n = 1}^N \frac{\ln n}{n^{1.1}} + \sum_{n > N} \frac{\ln n}{n^{1.1}}.$$
The first sum is finite. The second sum is bounded above by
$$ \sum_{n > N} \frac{n^{0.05}}{n^{1.1}},$$
which converges straightforwardly.
More generally, you can show that
$$ \sum_{n \geq 1} \frac{\ln n}{n^{\beta}}$$
converges absolutely for any $\beta > 1$. $\diamondsuit$
A: One has, by the integral convergence test and an integration by parts:
$$
0<\sum^{\infty}_{n=1}\frac{\ln{n}}{n^{1.1}}\leq \int^{\infty}_{1}\frac{\ln{x}}{x^{1.1}}dx=\left[-\frac{\ln x}{x^{0.1}}\right]^{\infty}_{1}+\int^{\infty}_{1}\frac{dx}{x^{1.1}}=100
$$ the series is convergent.
