Analysis of convergence of $\sum \frac{1}{\log^a n}$ I Need to show that
$$ \sum \frac{1}{(\log n)^a}$$
Diverges for all positive values of a.
My idea is to use the comparasion test, since the ratio test and rpot test are inconclusive. I want tp show that
$$ n^a > ln(n)$$ for values of n bigger than one $n_o$. This looks like a good approach, but i coulsnt find the $n_o$
Thanks in advance
 A: hint: compare with Harmonic series, which diverges
EDIT: consider the function $\frac{\log^k n}{n}$, set $\log n =t, \ n = e^t$. We need to show $a_n = \frac{e^n}{n^k}$ is an increasing function for some $n(k)$. Consider the ratio $\frac{a_{n+1}}{a_n}$. After a bit of algebra you can show it is larger than $1$ for $n >\frac{e^{-\frac{1}{k}}}{1-e^{-\frac{1}{k}}}$. 
Hence, $e^n >n^k \ \forall \ n> \frac{e^{-\frac{1}{k}}}{1-e^{-\frac{1}{k}}} \Rightarrow n >\log^k n \Rightarrow \frac{1}{n}<\frac{1}{\log^k n}$
A: You don't need to find the exact $n_0$ if you use l'Hopital's rule. Suppose you want to take $a = 1$. Then by l'Hopital
$$\lim_{n \rightarrow \infty} {\ln n \over n} = \lim_{n \rightarrow \infty } {{1 \over n} \over 1} = 0$$
So there will be some $n_0$ for which ${\displaystyle {\ln n \over n} < 1}$ for all $n > n_0$, which is all you need here. 
A: If you use the integral test, a substitution of $u=\ln(x)$ changes the integral to the equivalent: $$\int_{\ln(2)}^\infty \frac{e^{u}}{u^a} du.$$ l'Hopital's rule tells us that the integrand blows up to infinity for any $a$ and so the integral diverges.
Now recall that the integral test tells us for a decreasing continuous function $f(x)$, the series $\sum_{n=a}^\infty f(n)$ diverges iff the integral $\int_a^\infty f(x) dx$ diverges.
Thus we can conclude that the series diverges.
A: Since
$$
\lim_nn^{-s}(\ln n)^a=0 \quad \forall s\in (0,1],
$$
there is some $k\in \mathbb{N}$ such that 
$$
(\ln n)^a \le n^s \quad \forall n \ge k.
$$
There is no loss of generality in assuming that $k \ge 3$.
Then we have:
$$
\sum_{n=2}^\infty\frac{1}{(\ln n)^a}=\sum_{n=2}^{k-1}\frac{1}{(\ln n)^a}+\sum_{n=k}^\infty\frac{1}{(\ln n)^a}
\ge\sum_{n=2}^{k-1}\frac{1}{(\ln n)^a}+\sum_{n=k}^\infty n^{-s}
\ge\sum_{n=k}^\infty n^{-s}.
$$
Since the series $\sum_{n=k}^\infty n^{-s}$ diverges for $s\in (0,1]$, then 
the series $\sum_{n=2}^\infty\frac{1}{(\ln n)^a}$ also diverges.
