Is $\sum_{k=4}^{\infty }{k^{\log(k)}}/{(\log(k))^{k}}$ convergent or divergent? I came across this problem in a textbook, and the question is to investigate the convergence/divergence of the following series: $$\sum_{k=4}^{\infty }\frac{k^{\log(k)}}{(\log(k))^{k}}$$. I have no idea how to start solving this problem. I tried to call $a_{k}=\frac{k^{\log(k)}}{(\log(k))^{k}}$ and then proving that this limit maybe doesn't tend to zero and hence by the n-th term test the series diverges, but I couldn't do it. Any help is appreciated!!
 A: First hint:
$$
\left(\frac{k^{\log k}}{(\log k)^k}\right)^{1/k} = \frac{k^{\frac{\log k}{k}}}{\log k}.
$$
Second hint:
$$
\lim_{k \to \infty} k^{\frac{\log k}{k}} = 1. \qquad \text{(why?)}
$$
A: Another straightforward way is through the comparison test.  First of all, to 'normalize' and make comparing easier, let's rewrite both numerator and denominator in terms of a constant base: for the numerator we have $k^{\log k} = \left(e^{\log k}\right)^{\log k} = e^{(\log k)^2}$.  Likewise, for the denominator we have $(\log k)^k = \left(e^{\log\log k}\right)^k = e^{k\log\log k}$; so the overall expression is $\dfrac{e^{(\log k)^2}}{e^{k\log\log k}} = e^{(\log k)^2-k\log\log k}$.  Now, you can easily bound $\log\log k$ from below (for instance, it's $\gt 1$ as soon as $k\gt e^e$), and comparing rate of growth between the polylogarithmic term $(\log k)^2$ and the polynomial term $k$ (and this comparison is one to be heuristically familiar with, because it will come up a lot) should be enough to tell you how the series goes; can you see how you might make it rigorous from here?
