Why is $$\sum \frac{1}{\ln(n)^{\ln n}}$$ convergent?

I tried Cauchy's condensation test along with other test but nothing works well on this series. Any ideas?

  • 2
    $\begingroup$ Hint: $(\log(n))^{\log(n)} = n^{\log(\log(n))}$ . $\endgroup$ Jan 20 '18 at 18:09

$\log(n)$ is unbounded, in particular for $N$ big enough $\log(n) > e^2$ for all $n\ge N$. Then

$$ \sum_{n=N}^\infty \frac{1}{\log(n)^{\log(n)}} < \sum_{n=N}^\infty \frac{1}{e^{2\log(n)}} = \sum_{n=N}^\infty \frac{1}{n^2}$$

which converges, as you probably know.


Rewrite the general term as $$\frac1{(\ln n)^{\ln n}}=\frac1{e^{\ln n\,\ln(\ln n)}}=\frac1{n^{\ln(\ln n)}}.$$ Now $\dfrac 1{n^{\scriptstyle \ln(\ln n)}}=o\Bigl(\dfrac1{n^r}\Bigr)$ for any $r>0$. In particular it is $\;o\Bigl(\dfrac1{n^2}\Bigr)$, which converges.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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