# Is the following series converging $\sum_{n=2}\dfrac{1}{(\ln n)^{\ln n}}$ [duplicate]

Is the following series converging $\sum_{n=2}\dfrac{1}{(\ln n)^{\ln n}}$

I am not able to compare this with anything, can some show the way

## marked as duplicate by Brian Fitzpatrick, Jonas Meyer, saz, Hakim, Adam HughesOct 28 '14 at 19:09

Hint: Our denominator is $\exp((\ln\ln n)(\ln n))$, which is $n^{\ln\ln n}$. Now think $p$-series and Comparison.

Remark: When we meet $a^b$, the fact that it is equal to $\exp((\ln a)(b))$ is frequently useful.

• Ok thinking like $p$- series we see that $ln ln n >1$ so can we conclude directly that the series diverges? Why we need comparison? – Kushal Bhuyan May 3 '16 at 2:23
• Basically that's it, though early on $\ln\ln n\lt 1$. The term $p$-series is reserved for series $\sum\frac{1}{n^p}$ where $p$ is constant so ours is not a $p$-series, but after a while the $n$-th term is smaller than $\frac{1}{n^2}$ and we do comparison. – André Nicolas May 3 '16 at 3:42
• Hmmm that's right. – Kushal Bhuyan May 3 '16 at 4:39

HINT:
L'Hospitals rule gives us$$\lim_{n\to \infty}\dfrac{\ln n}{n}=0.$$ Therefore $$2\le\ln n\le n$$ for large $n$ values. $$\dfrac{1}{\ln n^{\ln n}}\le\dfrac{1}{2^n}.$$ I'm sure you can continue from here.

• I don't see how you got your last inequality from the preceding one. (Doesn't this imply that $n\ln 2\le (\ln n)(\ln(\ln n))$ for n large?) – user84413 Oct 29 '14 at 14:13
• @user84413: $$\dfrac{1}{\ln n}\le\dfrac{1}{2}$$ $$(\dfrac{1}{\ln n})^{\ln n}\le(\dfrac{1}{2})^n$$ – Bumblebee Oct 30 '14 at 13:59
• Thanks for your reply. I'm not sure this works, though. For example, $1/3\le1/2$, but $(\frac{1}{3})^2>(\frac{1}{2})^4$. – user84413 Oct 30 '14 at 15:06
• @user84413: I see. Since $\frac{1}{2}<1,$ my second inequality is not true. I have to find a better way to do this. Thank you. – Bumblebee Oct 31 '14 at 3:26

If $n\ge e^{(e^2)}$, then $\ln n\ge e^2\implies \ln(\ln n)\ge 2\implies$

$(\ln n)(\ln(\ln n))\ge 2 \ln n\implies \ln[(\ln n)^{\ln n}]\ge\ln(n^2)\implies(\ln n)^{\ln n}\ge n^2\implies \frac{1}{(\ln n)^{\ln n}}\le\frac{1}{n^2}$,

so the series converges by the Comparison Test with $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n^2}$.