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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!!

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    $\begingroup$ I imagine you want to use the root test. $\endgroup$ – James Jun 16 '13 at 19:59
  • $\begingroup$ Starting at $k=4$ seems really random. $\endgroup$ – Git Gud Jun 16 '13 at 20:00
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    $\begingroup$ What is $\log$ of the $k$th term? $\endgroup$ – Ted Shifrin Jun 16 '13 at 20:03
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First hint:

$$ \left(\frac{k^{\log k}}{(\log k)^k}\right)^{1/k} = \frac{k^{\frac{\log k}{k}}}{\log k}. $$

Second hint:

$$ \lim_{k \to \infty} k^{\frac{\log k}{k}} = 1. \qquad \text{(why?)} $$

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  • $\begingroup$ I understand both hints. Indeed, if we call $b_{k}=k^{\frac{logk}{k}}$, then $log(b_{k})=\frac{(logk)^2}{k}$ which tends to zero as $k$ tends to infinity, and hence $b_{k}$ tends to $1$. So, $\lim_{k\to \infty }\frac{k^{\frac{logk}{k}}}{logk}=0< 1$, which implies convergence by the root test. IS that correct? $\endgroup$ – Alex K Jun 16 '13 at 20:17
  • $\begingroup$ @AlexK, Yep! ${}$ $\endgroup$ – Antonio Vargas Jun 16 '13 at 20:20
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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?

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