3

I wanted to test the convergence of the series $$\sum_{n=3}^{\infty}\frac{1}{(\log\log n)^{\log n}}.$$ First I, apply the Cauchy condensation test i.e., $\displaystyle\sum_{n=3}^{\infty}\frac{2^n}{(\log\log 2^n)^{\log2^n}}$ which is same as $\displaystyle\sum_{n=3}^{\infty}\frac{2^n}{[\log(n\log2)]^{n\log 2}}$.

Next apply, root test i.e., $\displaystyle \lim_{n\to \infty}\left[\frac{2^n}{[\log(n\log2)]^{n\log 2}}\right]^{1/n}$ which becomes $\displaystyle \lim_{n\to \infty}\frac{2}{[\log(n\log2)]^{\log 2}}=0<1$, by root test the series $\displaystyle\sum_{n=3}^{\infty}\frac{2^n}{[\log(n\log2)]^{n\log 2}}$, converges, hence Cauchy Condensation test the series $\displaystyle\sum_{n=3}^{\infty}\frac{1}{(\log\log n)^{\log n}}$ converges.

Is this correct? or any other simple technique available?

Angelo
  • 12,328
Bhukya
  • 57
  • 2
    This seems correct yes, though you'd need to show the decreasingness of the general term (there are counterexamples to the test when this is not satisfied). Another approach is seeing that $1/(\log\log n)^{\log n} = 1/n^{\log\log\log n}$, and observe that for $n \geq n_0$ for some "big enough" $n_0$, $\log\log\log n \geq 2$ for example, meaning that for those $n$ you can compare the series with the one of $1/n^2$ which converges. – Bruno B Nov 11 '22 at 18:09
  • 2
    @BrunoB, I think that $,n_0=10^{703}$ is big enough as long as the function $,\log,$ is the natural logarithm. – Angelo Nov 11 '22 at 18:40
  • 1
    $n_0=4000000$ is more apprehensible and we still get $\frac 1{n^\alpha}$ with $\alpha>1$ (just slightly). – zwim Nov 11 '22 at 20:50

0 Answers0