Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Define $$S_n (a, b)=\sum_{k=a}^{b} {\frac{1}{k(\log k)(\log \log k)\cdots(\log^n k)}}$$

where $\log^n$ denotes $n$ compositions of the natural logarithm. And $$P(n>2)=\lfloor \underbrace{\exp(\exp(\cdots\exp}_{n-2\text{ times}}(1)\cdots)) \rfloor+1$$ so that $P(0)=1$, $P(1)=2$, $P(2) = 2$, $P(3) = \lfloor e \rfloor +1 = 3$, and so on. The idea is that $P(n)$ is the lowest integer that we can call $S_n(P(n), b)$ and have the logarithms output real numbers.

I know that $S_n (P(n), \infty)$ diverges arbitrarily slowly for large $n$. Then I suspect that

$$\lim_{n \to \infty} S_n(P(n), \infty)$$ diverges. It seems that this divergence depends on the definition of $S_n(a, \infty)$, which I take to be $\lim \limits_{b \to \infty} S_n(a,b)$. I am not sure how to verify/say this, and I am interested in someone else's take on this.

I was also wondering about the behavior of:

$$\lim_{n \to \infty} S_n(P(n), P(n+t))$$ for $t>0$. It seems to me that this sum should converge (to $0$) since it converges for arbitrary $n$, but at the same time it approaches $\lim \limits_{n \to \infty} S_n(P(n), \infty)$, which I suspect diverges.

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.