This is close to the ratio of arithmetic to geometric means for logarithms of natural numbers (here $\log$ is the natural logarithm):

$$\frac{\log 2+\log 3+\dots + \log n}{n \sqrt[n]{\log 2 \cdot \log 3 \cdots \log n}}=\frac{\log (n!)}{n \sqrt[n]{\log 2 \cdot \log 3 \cdots \log n}}$$

Numerically, the limit seems to be approaching $1$ from above, however the convergence is slow enough that I have doubts:

enter image description here

I do not know how to prove this limit, since the denominator doesn't have any nice closed form. I could use the Stirling approximation for the numerator, but what good would it do?

How can we prove the value of this limit (if it's even correct)?

$$\lim_{n \to \infty} \frac{\log (n!)}{n \sqrt[n]{\log 2 \cdot \log 3 \cdots \log n}}=1$$

Additionally, it's quite interesting to me that the sequence has a maximum at $n=35$ which looks like this:

enter image description here

As an additional question, I would like to know how to prove this maximal value (at least the fact that the sequence has only one maximum, which we then could find by simple evaluation for small $n$).

If we consider another sequence (with proper means this time, since there's $n-1$ logarithms for the $n$th element):

$$\frac{\log (n!)}{(n-1) \sqrt[n-1]{\log 2 \cdot \log 3 \cdots \log n}}$$

It shows the same behaviour but with a maximum for $n=12$.

One idea is to take logarithm of the expression:

$$\log \left(\frac{\log (n!)}{n \sqrt[n]{\log 2 \cdot \log 3 \cdots \log n}} \right)=\log \left(\log (n!) \right)- \log n -\frac{1}{n} \left( \log (\log 2)+ \dots +\log (\log n) \right)$$

I suppose this is easier to work with though I still have no idea how to prove the limit.

  • $\begingroup$ Use Stirling, change your sequence $f$ into $log(f)$, remember that $log(ab)=log(a)+log(b)$ and $log(\frac{a}{b})=log(a)-log(b)$ and finally that the log of the root can be expressed as $e^{\frac{log\big(log(2)\ldots log(n)\big)}{n}}$ should help you for the first part. $\endgroup$ Jan 30, 2018 at 1:12
  • 1
    $\begingroup$ You should be able to use Euler-Maclaurin for $\sum_{i\leq n}\log\log i$, noting that the (indefinite) integral is $x\log\log x-li(x)$; meanwhile $\log\log(n!)\approx\log(n(\log n-1))\approx\log n\log\log n$ - and yes, the next-order terms in all of these matter a lot, but it at least proides an angle of approach... $\endgroup$ Jan 30, 2018 at 1:14

1 Answer 1

  • Notice that if $f : [a, \infty) \to \mathbb{R}$ is increasing, then

    $$ \int_{m-1}^{n} f(x) \, dx \leq \sum_{k=m}^{n} f(k) \leq \int_{m}^{n+1} f(x) \, dx. $$

  • Applying the previous observation to $f(x) = \log x$, we have

    \begin{align*} \sum_{k=1}^{n} \log k &= \int_{1}^{n} \log x \, dx + \mathcal{O}(\log n) \\ &= n \log n - n + \mathcal{O}(\log n) \end{align*}

    where the last line follows from integration by parts.

  • Similarly, applying the previous observation to $f(x) = \log\log x$ we have

    \begin{align*} \sum_{k=2}^{n} \log\log k &= \int_{2}^{n} \log\log x \, dx + \mathcal{O}(\log\log n) \\ &= \left[ x \log\log x \right]_{x=2}^{x=n} - \int_{2}^{n} \frac{dx}{\log x} \\ &= n \log\log n + \mathcal{O}\left(\frac{n}{\log n}\right) \end{align*}

    where the last line follows from the L'Hospital's rule:

    $$ \frac{\int_{2}^{n} dx / \log x}{n / \log n} \sim \frac{\frac{1}{\log n}}{\frac{1}{\log n} - \frac{1}{\log^2 n}} \sim 1. $$

  • Therefore we have

    $$ \frac{\frac{1}{n}\sum_{k=1}^{n}\log k}{\exp\left(\frac{1}{n}\sum_{k=2}^{n}\log\log k \right)} = \frac{\log n - 1 + o(1)}{\exp(\log\log n + o(1))} = 1 + o(1)$$

    as desired. In fact, the computation tells that

    $$ \frac{\frac{1}{n}\sum_{k=1}^{n}\log k}{\exp\left(\frac{1}{n}\sum_{k=2}^{n}\log\log k \right)} = 1 + \mathcal{O}\left(\frac{1}{\log n}\right)$$

    and I believe that this bound is sharp up to generic constant.

  • 1
    $\begingroup$ That was fast. I shouldn't be surprised. To be honest, before Steven Stadnicki's comment, it didn't occur to me to use integrals, you further clarified the method to me, thank you $\endgroup$
    – Yuriy S
    Jan 30, 2018 at 1:38
  • $\begingroup$ @YuriyS, Glad it helped. :) I am pretty sure that you could be able to devise your own solution if you recognized an appropriate line of attack. Anyway, as Steven Stadnicki pointed out, Euler-MacLaurin-type argument seems to work very nicely when you are summing 'slowly increasing functions'. $\endgroup$ Jan 30, 2018 at 1:44

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