Sum of $ \ \frac{1}{(\ln k)^{\ln k}} \ $ How do I find out if the infinite sum of $ \ \frac{1}{(\ln k)^{\ln k}} \ $ is convergent or divergent? I'm given a hint: $ \ \ln k \ = \ e^{\ln(\ln k)}$ but I can't figure out how to apply that.
 A: For $k$ large enough you have $\ln k > e^2$, so the corresponding term ${\displaystyle {1 \over (\ln k)^{\ln k}}}$ satisfies
$$ {1 \over (\ln k)^{\ln k}}  < {1 \over e^{2\ln k}} = {1 \over k^2}$$
So the series converges by comparison with the sum of ${\displaystyle {1 \over k^2}}$. 
A: Using the hint:
$$S(k) = \frac{1}{(ln(k))^{ln(k)}} = \frac{1}{(e^{ln(ln(k))})^{ln(k)}} = \frac{1}{e^{ln(k)\cdot ln(ln(k))}} = \frac{1}{k^{ln(ln(k))}}$$
For $k>1$ it has only positive values. Note that:
$$ k \geq 16 \Rightarrow ln(ln(k)) \geq ln(ln(16)) > 1 \Rightarrow S(k) \leq \frac{1}{k^{ln(ln(16))}}$$ so the series converges by direct comparison with a p-series having $p = ln(ln(16)) > 1$.
See https://www.math.hmc.edu/calculus/tutorials/convergence/ for some info on the p-series and the comparison test.
A: HINT:
Write the inequality
$$\sum_{k=4}^{\infty}\frac{1}{(\log(k))^{\log k}}<\int_3^{\infty}\frac{1}{(\log(x))^{\log x}}\,dx=\int_{\log 3}^{\infty}\frac{1}{(x/e)^x}\,dx$$
then analyze the convergence of the resulting integral.
