Convergence of $\sum_{n=2}^{\infty} (\frac{\ln n}{n})^a$ depending on $a$ Calculate the convergence of $\sum_{n=2}^{\infty} (\frac{\ln n}{n})^a$ depending on the values of $a \in\mathbb R$.
If a = 1, then $\sum_{n=2}^{\infty} (\frac{\ln n}{n})^1 > \sum_{n=2}^{\infty} (\frac{1}{n})$, so it is divergent.
What should I do in case $a < 1$ or $a > 1$?
Edit: Can I say for a < 0 $\sum_{n=2}^{\infty} (\frac{\ln n}{n})^a $ = $\sum_{n=2}^{\infty} (\frac{n}{\ln n})^{(-a)}  $  = $\sum_{n=2}^{\infty} (\frac{n}{\ln n})^{b} $ (where b = -a and b > 0 )
$\sum_{n=2}^{\infty} (\frac{n}{\ln n})^{b}  $ = $\sum_{n=2}^{\infty} (\frac{n^b}{b\ln n})  $ > $\sum_{n=2}^{\infty} (\frac{1}{b\ln n})  >  \sum_{n=2}^{\infty} (\frac{1}{bn})  $, so it is divergent.
and for a > 0 $\sum_{n=2}^{\infty} (\frac{\ln n}{n})^a$ = $\sum_{n=2}^{\infty} (\frac{a\ln n}{n^a})$ < $\sum_{n=2}^{\infty} (\frac{a n}{n^a})$ = $\sum_{n=2}^{\infty} (\frac{a}{n^{a-1}})$, so for a -1 > 1 => a>2 it is convergent. And for 0<a<2 is it divergent
Is this correct?
 A: If $a\leq 0$ then $\left(\dfrac{\ln(n)}{n}\right)^a$ doesn't tend to $0$ as $n\rightarrow\infty$ and the series is obviously divergent. Your argument is correct but the series is so far from being convergent that is is unnecessarily complicated.
If $0<a<1$ we have since $\dfrac{\ln(n)}{n}<1$ (is that fact obvious enough?)
$$\left(\dfrac{\ln(n)}{n}\right)^a>\dfrac{\ln(n)}{n}$$
The RHS is divergent by your analysis so the LHS is divergent as well by a comparison test.
If $a>1$, let $b$ be any number such that $1<b<a$. Then:
$$\dfrac{\left(\dfrac{\ln(n)}{n}\right)^a}{\dfrac{1}{n^b}}=\dfrac{\ln(n)^a}{n^{a-b}}$$
Since $a-b>0$ we obtain by standard results on limit comparisons that
$$\dfrac{\left(\dfrac{\ln(n)}{n}\right)^a}{\dfrac{1}{n^b}}\rightarrow 0$$
as $n\rightarrow\infty$. Moreover the series
$$\sum_{n=2}^{\infty}\dfrac{1}{n^b}$$
is convergent since $b>1$. It follows from comparison test on series that
$$\sum_{n=2}^{\infty}\left(\dfrac{\ln(n)}{n}\right)^a$$
is convergent as well - its summand is dominated by a convergent series.
