I was looking at a list of primes. I noticed that $ \frac{AM (p_1, p_2, \ldots, p_n)}{p_n}$ seemed to converge.

This led me to try $ \frac{GM (p_1, p_2, \ldots, p_n)}{p_n}$ which also seemed to converge.

I did a quick Excel graph and regression and found the former seemed to converge to $\frac{1}{2}$ and latter to $\frac{1}{e}$. As with anything related to primes, no easy reasoning seemed to point to those results (however, for all natural numbers it was trivial to show that the former asymptotically tended to $\frac{1}{2}$).

Are these observations correct and are there any proofs towards:

$$ { \lim_{n\to\infty} \left( \frac{AM (p_1, p_2, \ldots, p_n)}{p_n} \right) = \frac{1}{2} \tag1 } $$

$$ { \lim_{n\to\infty} \left( \frac{GM (p_1, p_2, \ldots, p_n)}{p_n} \right) = \frac{1}{e} \tag2 } $$

Also, does the limit $$ { \lim_{n\to\infty} \left( \frac{HM (p_1, p_2, \ldots, p_n)}{p_n} \right) \tag3 } $$ exist?

  • 1
    $\begingroup$ Can we take $p_n = n\cdot log(n)$ in these limits? (using the prime numbers theorem) $\endgroup$ – Yanko Sep 15 '18 at 14:45
  • 1
    $\begingroup$ What are AM, GM & HM? $\endgroup$ – corey979 Sep 15 '18 at 17:11
  • 7
    $\begingroup$ @corey979 Arithmetic, geometric, and harmonic means; together, these are known as the Pythagorean means. $\endgroup$ – Théophile Sep 15 '18 at 17:18
  • 9
    $\begingroup$ For what it's worth, the answer to the question in the title is "No". $\endgroup$ – Eric Duminil Sep 15 '18 at 17:37
  • 5
    $\begingroup$ @Soham Do not edit your question to ask more questions. It invalidates existing answers. Ask a new question instead. $\endgroup$ – user202729 Sep 16 '18 at 2:48

Your conjecture for GM was proved in 2011 in the short paper On a limit involving the product of prime numbers by József Sándor and Antoine Verroken.

Abstract. Let $p_k$ denote the $k$th prime number. The aim of this note is to prove that the limit of the sequence $(p_n / \sqrt[n]{p_1 \cdots p_n})$ is $e$.

The authors obtain the result based on the prime number theorem, i.e., $$p_n \approx n \log n \quad \textrm{as} \ n \to \infty$$ as well as an inequality with Chebyshev's function $$\theta(x) = \sum_{p \le x}\log p$$ where $p$ are primes less than $x$.

  • 5
    $\begingroup$ @TheSimpliFire I think we simply can't conclude that the sequence is strictly decreasing based on the first $15$ values. In a paper from 2016 on the same topic, the author points out that the sequence converges slowly. $\endgroup$ – Théophile Sep 15 '18 at 15:01
  • 2
    $\begingroup$ @TheSimpliFire Also, going to a third digit clarifies the issue: the $GM/p_n$ for $n=5$ is about $0.428$ and the $GM/p_n$ for $n=6$ is about $0.429$, so even in the beginning the sequence is not strictly decreasing. $\endgroup$ – Mark S. Sep 15 '18 at 17:37
  • 2
    $\begingroup$ @Soham Good question. For positive $x_i$, we have $\min(x_1,\ldots,x_n) \le HM(x_1,\ldots,x_n) \le n\min(x_1,\ldots,x_n)$. This means that for the first $n$ primes, $$2 \le HM(p_1,\ldots,p_n) \le 4,$$ so the limit of $HM/p_n$ will vanish to $0$. $\endgroup$ – Théophile Sep 15 '18 at 17:43
  • 1
    $\begingroup$ @Théophile Do you mean $2 \leq HM(p_1, \ldots, p_m) \leq 2n$? The result still follows, of course, since $\frac{2n}{n\log n} \to 0$. $\endgroup$ – mathmandan Sep 15 '18 at 17:56
  • 1
    $\begingroup$ @TheSimpliFire : the formula in the cited literature is just your reciprocal. Additionally, if you use $p_{n-1}$ in the denominator of your formula it looks as if this could as well converge to $e$... $\endgroup$ – Gottfried Helms Sep 15 '18 at 18:40

We can use the simple version of the prime counting function $$p_n \approx n \log n$$ and plug it into your expressions. For the arithmetic one, this becomes $$\lim_{n \to \infty} \frac {\sum_{i=1}^n p_i}{np_n}=\lim_{n \to \infty} \frac {\sum_{i=1}^n i\log(i)}{np_n}=\lim_{n \to \infty} \frac {\sum_{i=1}^n \log(i^i)}{np_n}\\=\lim_{n \to \infty} \frac {\log\prod_{i=1}^n i^i}{np_n}=\lim_{n \to \infty}\frac {\log(H(n))}{n^2\log(n)}$$ Where $H(n)$ is the hyperfactorial function. We can use the expansion given on the Mathworld page to get $$\log H(n)\approx \log A -\frac {n^2}4+\left(\frac {n(n+1)}2+\frac 1{12}\right)\log (n)$$ and the limit is duly $\frac 12$

I didn't find a nice expression for the product of the primes.

  • 2
    $\begingroup$ Thanks. Not as though I understand the proof, though..haha $\endgroup$ – Soham Sep 15 '18 at 16:39
  • $\begingroup$ So, I just tried the same with HM too..This one I'm not sure, converged or not, to a non-zero number..could well be going to zero..For the first 1 lakh numbers, it dropped down to 0.02 levels..Does this limit exist? $\endgroup$ – Soham Sep 15 '18 at 16:46
  • $\begingroup$ I added some detail of how I got to $H(n)$. The asymptotic expansion came from the Mathworld page I linked to. $\endgroup$ – Ross Millikan Sep 15 '18 at 22:59
  • 2
    $\begingroup$ In which sense are you using the $\approx$ symbol? It's not clear to me that you can simply replace $p_i$ with $i \log i$ in your computations. In general this kind of replacement properties do not hold. $\endgroup$ – Federico Poloni Sep 17 '18 at 11:46
  • 3
    $\begingroup$ @FedericoPoloni: It is valid with the standard definition, $p_n \approx n \log(n) \iff p_n = n \log(n) + o(p_n)$. If you plug this exact value into $\frac{\sum p_i}{np_n}$ you get $\frac{\sum i\log(i)}{np_n} + \frac{o(p_n)}{np_n}$. The rightmost part vanishes to $0$ near infinity so everything is fine. $\endgroup$ – Mariuslp Sep 18 '18 at 13:02

Here is a general answer to this which will solve the case for AM, GM and HM in one shot.

Observe that since $p_n \sim n\log n$, as $n \to \infty$ the proportion of numbers formed by the sequence of ratios $\frac{p_1}{p_n},\frac{p_2}{p_n} \ldots, \frac{p_{n-1}}{p_n}$ which fall inside any sub-interval within $(0,1)$ is proportion to the length of that interval i.e. the sequence $\frac{p_r}{p_n}$ approaches equidistributed in $(0,1)$ [for the proof of equi/uniform distribution, see the comment below by Mariuslp]. Hence, for an equidistributes sequence, we have:

Theorem: Let $p_k$ be the $k$-th prime and let $f$ be a continuous function Riemann integrable in $(0,1)$ then,

$$ \lim_{n \to \infty}\frac{1}{n}\sum_{r = 1}^{n}f\Big(\frac{p_r}{p_n}\Big) = \int_{0}^{1}f(x)dx. $$

(See the proof in this MO link for a direct proof). Taking $f(x) = x, \log(x)$ and $\frac{1}{x}$ respectively with some manipulations, we get the required limit for AM, GM and HM as $\frac{1}{2},\frac{1}{e}$ and $0$ respectively.

Example: Showing the case for GM due to request in the bellow comments. Let $$ \lim_{n \to \infty}\frac{(p_1 p_2 \ldots p_n)^{1/n}}{p_n} = \lim_{n \to \infty}\Big(\frac{p_1}{p_n}\Big)^{1/n} \Big(\frac{p_2}{p_n}\Big)^{1/n} \ldots \Big(\frac{p_n}{p_n}\Big)^{1/n} = l $$

Clearly, $0 < l < 1$. Taking logarithm on both sides, we have $$ \log l = \lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n}\log \Big(\frac{p_r}{p_n}\Big) = \int_{0}^{1} \log x dx = -1. $$

Hence $l = 1/e$.

  • 12
    $\begingroup$ In which sense are those ratios uniformly distributed? What is the probability space here? And isn't the last of those ratios always 1? $\endgroup$ – Federico Poloni Sep 17 '18 at 6:46
  • 2
    $\begingroup$ @TorstenSchoeneberg: For the second case, the left hand side is the logarithm of the GM, so the right hand side needs to be $-1$, not $1/e$. Also, in the third case you've got 1/HM on the left. $\endgroup$ – celtschk Sep 17 '18 at 14:46
  • 4
    $\begingroup$ This is a great answer. $\endgroup$ – Pakk Sep 18 '18 at 6:28
  • 1
    $\begingroup$ @Mariuslp: There is no bias towards $0$. Take $n = 100000$ or any big number, compute the ratios and convince yourself experimentally. Alternatively , use the asymptotic expansion of the $n$-th prime to convince yourself theoretically. That's all I can say. $\endgroup$ – Nilotpal Kanti Sinha Sep 18 '18 at 13:21
  • 6
    $\begingroup$ @NilotpalKantiSinha: As counterintuitive as it is to me you are right. For those who struggled like me, in $[\alpha, \beta]$ there are $\pi(\beta p_n) - \pi(\alpha p_n) \approx \frac{\beta p_n}{\ln (\beta p_n)} - \frac{\alpha p_n}{\ln (\alpha p_n)}$ primes, on a total of $\pi(p_n) \approx \frac{p_n}{\ln(p_n)}$. Knowing that $\ln(an) \approx \ln(n)$ for any $a>0$, with some manipulations (remember we cannot replace by equivalents in a sum), we do get a $\beta - \alpha$ proportion of primes in the interval. $\endgroup$ – Mariuslp Sep 18 '18 at 14:24

Not an answer, but an illustration of the type of convergence of (2). I reproduced the formula for the geometrical mean in the version reciprocal to the OP's formula to match the formula of the cited literature. The curve shows the deviation from $e$ and the slowness of convergence.


The red curve is the running mean using 7 data points.


About the harmonic mean: I believe that it exists, but it is zero.

The harmonic mean limit (HML) is $$HML=\lim_{n\rightarrow\infty}\left(\frac{HM(p_1,p_2,\ldots)}{p_n}\right).$$ The harmonic mean itself can be written as $$HM(p_1,p_2,\ldots)=\frac{n}{\sum_{i=1}^{n}\frac{1}{p_i}}.$$ The asymptotic behavior of the sum in the fraction is (see Mathworld): $$\sum_{i=1}^{n}\frac{1}{p_i}=\ln \ln p_n + B_1 + o(1),$$ with $B_1 \approx0.261$ the Mertens constant, so the asymptotic behavior of the harmonic mean is $$HM(p_1,p_2,\ldots)=O\left(\frac{n}{\ln \ln p_n}\right)=o(n).$$

Here, $o(n)$ is the small oh-notation, which means that asymptotically, the left hand term is smaller than $n$.

Using the approximation $p_n\approx n\ln n$, $$HML=\lim_{n\rightarrow\infty}\left(\frac{o(n)}{n\ln n}\right)=\lim_{n\rightarrow\infty}o\left(\frac{1}{\ln n}\right)=0.$$

  • 1
    $\begingroup$ Personally I wouldn't bother with the last three "simplifications" before saying the limit is zero. The first expression is "obviously" $1/g(n)$ where $g(n)$ tends (slowly) to infinity as $n$ tends to infinity $\endgroup$ – Martin Bonner Sep 18 '18 at 7:18
  • $\begingroup$ @MartinBonner: You're right, and I think a few more steps can be removed... $\endgroup$ – Pakk Sep 18 '18 at 7:22

This is a lower-tech alternative to Ross Millikan's answer for the Arithmetic Mean result (not using the hyperactive factorial function...), followed by a proof for the Geometric Mean (which occurred to me later).

Using $p_n\approx n\log n$ for large $n$, we need to prove

$${1\over n^2}\sum_{k=1}^n{k\log k\over\log n}\to{1\over2}$$

But for any $0\lt r\lt1$ we have

$$(1-r)\sum_{k=\lceil n^{1-r}\rceil}^nk\le\sum_{k=1}^n{k\log k\over\log n}\le\sum_{k=1}^nk={n(n+1)\over2}$$

(Note, the lower limit on the lower bounding sum is $k=\lceil n^{1-r}\rceil$; for some reason it doesn't render well in the displayed version on my screen.) Now

$$\sum_{k=\lceil n^{1-r}\rceil}^nk={n(n+1)-(\lceil n^{1-r}\rceil-1)\lceil n^{1-r}\rceil\over2}\ge{n(n+1)-n^{1-r}(n^{1-r}+1)\over2}\ge{n^2\over2}\left(1-{1\over n^{2r}}-{1\over n^{1+r}} \right)\ge{n^2\over2}\left(1-{2\over n^{2r}} \right)$$

It follows that

$${1-r\over2}\left(1-{2\over n^{2r}}\right)\le{1\over n^2}\sum_{k=1}^n{k\log k\over\log n}\le{1\over2}\left(1+{1\over n}\right)$$

Finally, let's let $r=1/\sqrt{\log n}$ Then $n^{2r}=e^{2r\log n}=e^{2\sqrt{\log n}}\to\infty$ as $n\to\infty$. It follows that

$${1-r\over2}\left(1-{2\over n^{2r}}\right)={1-(1/\log n)\over2}\left(1-{2\over e^{2\sqrt{\log n}}}\right)\to{1\over2}$$

and the Squeeze Theorem does the rest.

Added later: The low-tech approach also handles the Geometric Mean. Since $p_n/(n\log n)\to1$ as $n\to\infty$, it suffices to show

$${1\over n}\sum_{k=1}^n\log\left(n\log n\over p_k\right)\to1$$

If we write $p_k=(1+\epsilon_k)k\log k$ (for $k\gt1$) and note that $\epsilon_k\to0$ as $k\to\infty$, and again let $r=1/\sqrt{\log n}$ (with $n\gt2$), we have

$${1\over n}\sum_{k=1}^n\log\left(n\log n\over p_k\right)={1\over n}\sum_{k=1}^{\lfloor n^{1-r}\rfloor}\log\left(n\log n\over p_k\right)+{1\over n}\sum_{k=\lceil n^{1-r}\rceil}^n\log\left(n\log n\over (1+\epsilon_k)k\log k\right)\\ ={1\over n}\sum_{k=1}^{\lfloor n^{1-r}\rfloor}\log\left(n\log n\over p_k\right) -{1\over n}\sum_{k=\lceil n^{1-r}\rceil}^n\log(1+\epsilon_k) +{1\over n}\sum_{k=\lceil n^{1-r}\rceil}^n\log\left(\log n\over\log k\right) -{1\over n}\sum_{k=\lceil n^{1-r}\rceil}^n\log\left(k\over n\right)$$

Now for large $n=e^{u^2}$ (so that $r=1/u$), we have

$$0\lt{1\over n}\sum_{k=1}^{\lfloor n^{1-r}\rfloor}\log\left(n\log n\over p_k\right)\lt{1\over n}\sum_{k=1}^{\lfloor n^{1-r}\rfloor}\log\left(n\log n\over 2\right)\lt{\log(n\log n/2)\over n^r}={u^2+2\log u-\log2\over e^u}\to0$$


$$0\lt{1\over n}\sum_{k=\lceil n^{1-r}\rceil}^n\log\left(\log n\over\log k\right)\lt{1\over n}\sum_{k=\lceil n^{1-r}\rceil}^n\log\left(\log n\over\log n^{1-r}\right)\lt\log(1-r)\to0$$

We also have

$$\left|{1\over n}\sum_{k=\lceil n^{1-r}\rceil}^n\log(1+\epsilon_k)\right|\le\max_{\lceil n^{1-r}\rceil\le k\le n}|\log(1+\epsilon_k)|\to0$$

since the range over which the max is taken forces $\epsilon_k\to0$. Finally, we have

$$-{1\over n}\sum_{k=\lceil n^{1-r}\rceil}^n\log\left(k\over n\right) = {1\over n}\sum_{k=1}^{\lfloor n^{1-r}\rfloor}\log\left(k\over n\right) -{1\over n}\sum_{k=1}^n\log\left(k\over n\right)$$


$$0\lt{1\over n}\sum_{k=1}^{\lfloor n^{1-r}\rfloor}\log\left(n\over k\right)\lt{\log n\over n^r}={u^2\over e^u}\to0$$


$$-{1\over n}\sum_{k=1}^n\log\left(k\over n\right)\to-\int_0^1\log x\,dx=1$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.