Let $p_n$ be the $n$-th prime and $F_n$ be the $n$-th Fibonacci number. We have

$$ \lim_{n \to \infty}\frac{(p_1 p_2 \ldots p_n)^{1/n}}{p_n} = \lim_{n \to \infty}\frac{\{\log(F_3)\log(F_4)\ldots \log(F_n)\}^{1/n}}{\log(F_n)} = \frac{1}{e} $$

The first limit was proved by Sandor and Verroken using the prime number theorem and the Chebyshev function. The second limit was proved by Farhadian and Jakimjuk using the Binet's formula for Fibonacci numbers and Stirling's approximation for factorial.

Although these two results were proved using different ingredients, their structure is exactly similar which led me to investigate if there is a stronger phenomenon governing such results. My analysis led me to the following.

Claim: If $a_n= n^{1+o(1)}$ is increasing then, $ \lim_{n \to \infty}\dfrac{(a_1 a_2\ldots a_n)^{1/n}}{a_n} = \dfrac{1}{e}.$

I believe that I have a proof for this using Weyl's Equidistribution Theorem. I am looking for a simpler or a more elementary proof of this. Also have I got the conditions right for this claim to hold?

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    $\begingroup$ You might be interested in Matiyasevich and Guy, A new formula for $\pi$, The American Mathematical Monthly, Vol. 93, No. 8 (Oct., 1986), pp. 631-635, jstor.org/stable/2322322 $\endgroup$ – Gerry Myerson Apr 16 '19 at 6:16
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    $\begingroup$ If $a_n$ is an arbitrary sequence, what is the purpose of having a function $f$ ? (And why would you require differentiability on a discrete domain ?) $\endgroup$ – Yves Daoust Apr 16 '19 at 7:45

The logarithm of the ratio is $\overline{\log f(a_n)}-\log f(a_n)$ where the overline denotes the arithmetic average from $1$ to $n$.

For $f(a_n)=n^{1+\epsilon}$, it tends to

$$-(1+\epsilon),$$ so that the condition $f(a_n)=O(n^{1+\epsilon})$ is not the right one.

Generally speaking, there is no reason that $\overline{x_n}-x_n$ tends to $-1$ for $x_n$ slowly growing. Take $x_n=1-\dfrac1n$, for example, and the claim is false.

If I am right, a correct and simpler claim could be

$$\log(a_n)\sim\log(n)\implies\frac{(a_1\cdot a_2\cdots a_n)^{1/n}}{a_n}\to\frac 1e$$ or in other terms


This is obviously true for $a_n=\log(n)$, but we have to prove that additive terms cannot introduce a finite bias.

Let us assume that

$$\log(a_n)=\log n+o(\log (n)).$$


$$\overline{\log(a_n)}-\log(a_n)=\frac{(n\log n-n)+o(n)+n\,o(\log(n))}n-\log(n)-o(\log(n)) \\=-1+o(\log(n))$$ and it is possible that the sequence diverges.

Anyway, I failed to find a suitable term. For instance, with

$$\log(a_n)=\log(n)+\log(\log(n)),$$ we have, by the Euler-Maclaurin summation formula

$$\sum_2^n{\log\log(k))}\sim\int_2^n\log(\log(x))\,dx+\frac12\log(\log(n))\sim n\log(\log(n))-\frac n{\log(n)}+\frac12\log(\log(n)),$$

so that the residue is

$$\frac{-\dfrac n{\log(n)}+\dfrac12\log(\log(n))+\cdots}{n},$$ which tends to zero.

  • $\begingroup$ @NilotpalKantiSinha: $n\to0$ ??? $\endgroup$ – Yves Daoust Apr 16 '19 at 8:20
  • $\begingroup$ @NilotpalKantiSinha IMO, writing $O(n^{1+\epsilon})$ for some $\epsilon\ge0$ is a technical mistake. It should be $O(n^{1+\epsilon(n)})$, with $\epsilon(n)\ge0$ and $\lim_{n\to\infty}\epsilon(n)=0$. But this does not exclude $\sqrt n$, for instance. $\endgroup$ – Yves Daoust Apr 16 '19 at 8:26
  • $\begingroup$ Yes, that was I had in mind for epsilon. $\endgroup$ – Nilotpal Kanti Sinha Apr 16 '19 at 8:29
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    $\begingroup$ @NilotpalKantiSinha: I suggest that you stop disregarding my answer and fix your question. As it stands, your claim is false, and not due to $\epsilon$. $\endgroup$ – Yves Daoust Apr 16 '19 at 9:00
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    $\begingroup$ @YvesDaoust You new formulation looks elegant!! $\endgroup$ – Nilotpal Kanti Sinha Apr 16 '19 at 10:31

Write $a_n = n e^{c_n}$. This $c_n$ is related to OP's $\epsilon_n$ by $c_n = \epsilon_n \log n$. Now

  • Monotonicity of $a_n$ translates to $c_n + \log n \geq c_{n-1} + \log(n-1)$, or by rearranging, it is equivalent to $c_n - c_{n-1} \geq \log\left(1-\frac{1}{n}\right)$. This is obviously true if $c_n$ itself is increasing, but not necessarily restricted to such cases.

  • $a_n = n^{1+o(1)}$ is equivalent to $c_n = o(\log n)$.

Now plugging this to the quantity of our interest,

$$ \frac{(a_1 \cdots a_n)^{1/n}}{a_n} = \frac{(n!)^{1/n}}{n} e^{d_n} $$

where $d_n = \frac{1}{n}\left(\sum_{k=1}^{n} c_k\right) - c_n$ is the difference between the Cesàro-mean of $c_n$ and $c_n$ itself. Since $(n!)^{1/n}/n \to 1/e$ as $n\to \infty$, the question boils down to asking the following question on $(c_n)$.

Reformulation: Let $(c_n)$ satisfy

  1. $c_n - c_{n-1} \geq \log\left(1-\frac{1}{n}\right)$, and
  2. $c_n = o(\log n)$.

Does the following limit always hold?

$$ \lim_{n\to\infty} d_n = \lim_{n\to\infty} \left( \frac{\sum_{k=1}^{n} c_k}{n} - c_n \right) \quad \stackrel{?}{=} \quad 0 $$

Here we discuss several counter-examples.

Counter-example 1. Let

$$c_1 = 0 \qquad \text{and} \qquad c_n = \lfloor \log_2 \log_2 n \rfloor \quad \text{for} \quad n \geq 2. $$

It is clear that $(c_n)$ satisfies both conditions for the reformulated problem. On the other hand, by the reverse triangle inequality and the monotonicity of $c_n$,

\begin{align*} \left|d_n - d_{n-1} \right| &\geq |c_n - c_{n-1}| - \frac{1}{n(n-1)}\sum_{k=1}^{n-1} c_k - \frac{c_n}{n} \\ &\geq |c_n - c_{n-1}| - \frac{c_{n-1} + c_n}{n}. \end{align*}

Since $c_n - c_{n-1} = 1$ for infinitely many $n$'s (i.e. along $n = 2^{2^m}$ for $m=0,1,2,\cdots$) and $c_n/n \to 0$, we have $\liminf_{n\to\infty} |d_n - d_{n-1}| \geq 1$ and therefore $d_n$ cannot converge.

Alternatively, we can give a more direct argument. Let $n = 2^{2^m} - 1$ for a non-negative integer $m$. By grouping terms with the same values,

\begin{align*} \sum_{k=1}^{2^{2^m}-1} c_k = \sum_{i=0}^{m-1} i (2^{2^{i+1}}-2^{2^i}) = (m-1) 2^{2^m} - \sum_{i=1}^{m-1} 2^{2^i}. \end{align*}

The double-exponential growth quickly tells that the last sum is nigligible compared to $2^{2^m}$:

$$ \frac{1}{2^{2^m}} \sum_{i=1}^{m-1} 2^{2^i} \leq \frac{m \sqrt{2^{2^m}}}{2^{2^m}} \xrightarrow[m\to\infty]{} 0. $$

Together with $c_{2^{2^m}-1} = m-1$, it follows that

\begin{align*} d_{2^{2^m}-1} = \frac{\sum_{k=1}^{2^{2^m}-1} c_k}{2^{2^m}-1} - c_{2^{2^m}-1} = (m-1)\frac{2^{2^m}}{2^{2^m}-1} + o(1) - (m-1) = o(1) \end{align*}

On the other hand, with $n = 2^{2^m}$, we have $c_{2^{2^m}} = m$ and hence

\begin{align*} d_{2^{2^m}} = \frac{\sum_{k=1}^{2^{2^m}} c_k}{2^{2^m}} - c_{2^{2^m}} = (m-1)\frac{2^{2^m}}{2^{2^m}-1} + o(1) - m = -1 + o(1). \end{align*}

Therefore both $0$ and $-1$ are subsequential limits of $(d_n)$ and it cannot converge as $n\to\infty$. (This computation also confirms that $\liminf_{n\to\infty} |d_n - d_{n-1}| \geq 1$, with the lower bound realized by the subsequence along $n = 2^{2^m}$ as $m\to\infty$.

Counter-example 2. This example shares a similar feature as in the previous example. Let

$$c_n = \lfloor \log_3 n \rfloor - \log_3 n.$$


$$ c_n - c_{n-1} = \begin{cases} 1 + \log_3\left(1-\frac{1}{n}\right), & \text{if $n = 3^m$ for some $m\in\mathbb{Z}$}; \\ \log_3\left(1-\frac{1}{n}\right), & \text{otherwise}. \end{cases}$$

In either cases, we have $c_n - c_{n-1} \geq \log\left(1-\frac{1}{n}\right)$, thanks to $\frac{1}{\log 3} < 1$. Since $-1 < c_n \leq 0$, it is obviously $o(\log n)$.

Now let us investigate the limit in the reformulated problem. When $n = 3^m$ for some $m \in \mathbb{Z}$, we have $c_n = c_{3^m} = 0$. Also, splitting the sum according to the value of $i = \lfloor \log k \rfloor$ gives

\begin{align*} d_{3^m} &= \frac{1}{3^m}\sum_{k=1}^{3^m-1} (\lfloor \log_3 k \rfloor - \log_3 k) \\ &= \frac{1}{3^m}\sum_{i=0}^{m-1}\sum_{j=0}^{2\cdot3^i-1} \left( i - \log_3\left(3^i + j\right)\right) \\ &= - \frac{1}{3^m}\sum_{i=0}^{m-1}\sum_{j=0}^{2\cdot3^i-1} \log_3\left(1 + \frac{j}{3^i}\right) \\ &= \sum_{i=0}^{m-1} \frac{2\cdot 3^i}{3^m} \left( - \sum_{j=0}^{2\cdot3^i-1} \log_3\left(1 + \frac{2j}{2\cdot 3^i}\right) \frac{1}{2\cdot 3^i} \right). \end{align*}

Notice that the last sum can be understood as the weighted average of the Riemann sums of the integral $-\int_{0}^{1} \log_3(1+2x) \, \mathrm{d}x \approx -0.589761 $. Since larger $i$ receives much more weight, it can be easily proved (or by appealing to abelian theorem for Nørlund means) that

$$ \lim_{m\to\infty} d_{3^m} = -\int_{0}^{1} \log_3(1+2x). $$

On the other hand, it is also easy to check that $c_{3^m-1} \to -1$ as $m\to\infty$, and so,

$$ \lim_{m\to\infty} d_{3^m - 1} = 1 - \int_{0}^{1} \log_3(1+2x). $$

Since $(d_n)$ has two different subsequential limits, it cannot converge.

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    $\begingroup$ Mh, if I am right, what you suggest is essentially $a_n=n\log n$, and $\log a_n=\log n+\log\log n$. From WA, the sum of $\log\log k$ is asymptotic to $n\log\log n-\text{Li }n$. After division by $n$ and subtraction of $\log\log n$, there is no residue. $\endgroup$ – Yves Daoust Apr 16 '19 at 10:20
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    $\begingroup$ @YvesDaoust Flooring function plays crucial role in this argument. $c_n$ is almost constant with very occasional jump at doubly exponential points, which is enough to drastically change the behavior of it's partial sums from that of mere $\log\log n$. $\endgroup$ – Sangchul Lee Apr 16 '19 at 10:26
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    $\begingroup$ @YvesDaoust I agree that the partial sums for both ones will have the same leading order. But here we are interested in the difference between running average and the $n$-th sum, which amounts to capturing the next-leading order of the partial sums. This definitely captures the difference of them. Most of all, do you see any issue in my proof above? $\endgroup$ – Sangchul Lee Apr 16 '19 at 10:38
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    $\begingroup$ @YvesDaoust If neither of mathematical proof nor numerical simulation convinces you, I am not sure what else I can do. $\endgroup$ – Sangchul Lee Apr 16 '19 at 11:02
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    $\begingroup$ Sorry! you are quite right! I deleted my silly comment. $\endgroup$ – John Bentin Apr 16 '19 at 16:07

I will denote $b_n:=\dfrac{(a_1 a_2\ldots a_n)^{1/n}}{a_n}$.

Set $a_n = 0.01n + 0.99\cdot 10^{\lfloor \log_{10} n\rfloor}$. Then $\frac{n}{10}<a_n\leq n$, and $a_n$ is increasing, so it satisfies the condition. We have $$ \lim\limits_{n\to\infty}\frac{(a_1\dots a_n)^{1/n}}{(a_1\dots a_{n+1})^{1/(n+1)}}=1, $$ since $$ \ln \frac{(a_1\dots a_n)^{1/n}}{(a_1\dots a_{n+1})^{1/(n+1)}} = \frac{1}{n(n+1)}\ln (a_1\dots a_n) - \frac{1}{n+1}\ln a_{n+1}, $$ which is bounded between $-\frac {\ln (n+1)}{n+1}$ and $\frac {\ln n}{n+1}$.

On the other hand, $\lim\limits_{k\to\infty}\frac{a_{10^k}}{a_{10^k-1}}=\lim\limits_{k\to\infty}\frac{10^k}{1.09\cdot 10^{k-1}-0.01}=\frac{1000}{109}$.

Assume now that $\lim\limits_{n\to\infty} b_n$ exists and is positive. Then $$ 1 = {\lim\limits_{k\to\infty}\frac{b_{10^k-1}}{b_{10^k}}} = \lim\limits_{k\to\infty} \left(\frac{(a_1\dots a_{10^k-1})^{1/(10^k-1)}}{(a_1\dots a_{10^k})^{10^{-k}}}\cdot \frac{a_{10^k}}{a_{10^k-1}}\right) = 1\cdot \frac{1000}{109}, $$ which is a contradiction.


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