2
$\begingroup$

This question already has an answer here:

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.

$\endgroup$

marked as duplicate by Hans Lundmark, Misha Lavrov, user99914, Yujie Zha, JonMark Perry Dec 10 '17 at 4:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ Please advise as to whether I have preserved or altered the meaning of your term by editing the code. $\endgroup$ – colormegone Aug 25 '15 at 2:43
  • $\begingroup$ It is the same, thanks! $\endgroup$ – Nawar Ismail Aug 25 '15 at 3:01
1
$\begingroup$

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 ${\displaystyle {1 \over k^2}}$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola Sep 10 '15 at 2:54

Not the answer you're looking for? Browse other questions tagged or ask your own question.