Why is $\zeta(2) = \frac{\pi^2}{6}$ almost equal to $\sqrt{e}$?

Experimenting a bit I also found $\zeta(\frac{8}{3}) \approx e^\frac{1}{4}$, $\zeta(\frac{31}{9}) \approx e^\frac{1}{8}$ and $\zeta(\frac{141}{23}) \approx e^\frac{1}{64}$. I also figured out that $\zeta(x)$ approaches $e^{2^{-x}}$ but I'm not sure that helps explain why these almost-equalities exist. How to quantify how surprising these almost-equalities are, and what is the explanation for them if any?

EDIT: There does seem to be a pattern here: $\log \zeta(n + (\frac{2}{3})^{n-1}) \approx 2^{-n}$ for $n = 1,2,3,4,...$. I think this formula explains the observations but where does it come from?

BONUS, since I've retagged this as a soft-question already: Is there any wrong but somehow plausible argument that two random integers are relatively prime with probability $\frac{1}{\sqrt{e}}$? I guess it would be like a Lucky Larry story.


closed as not a real question by Rasmus, lhf, Asaf Karagila, Henning Makholm, Carl Mummert Feb 11 '12 at 3:37

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    $\begingroup$ They're really not that close. $\endgroup$ – Qiaochu Yuan Feb 9 '12 at 7:26
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    $\begingroup$ I checked this also, and I found numerically that there are infinitely many real numbers between $\zeta(2)$ and $\sqrt{e}$. $\endgroup$ – yohBS Feb 9 '12 at 7:29
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    $\begingroup$ For every $x\gt1$ and $\varepsilon\gt0$ there exists integers $a$ and $b$ such that $|\zeta(x)-\mathrm e^{a/b}|\leqslant\varepsilon$. Hence my quantification of the degree of surprise of the properties (not identities) you suggest is: NULL. // The asymptotics $\zeta(x)\approx\mathrm e^{2^{-x}}$ when $x\to+\infty$ is only natural since $\zeta(x)=1+2^{-x}+o(2^{-x})$ and $\mathrm e^{u}=1+u+o(u)$ when $u\to0$. $\endgroup$ – Did Feb 9 '12 at 7:36
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    $\begingroup$ Since you seem to have missed it the first time, let me repeat: $\zeta(n+\varepsilon_n)=1+2^{-n}+o(2^{-n})=\exp(2^{-n}+o(2^{-n}))$ when $n\to\infty$, for every $\varepsilon_n\to0$ (and $\zeta(n)=1+2^{-n}+O(3^{-n})=\exp(2^{-n}(1+O(a^{n}))$ with $a=2/3\lt1$). $\endgroup$ – Did Feb 9 '12 at 9:27
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    $\begingroup$ 1.64872... and 1.64493... are not that close at all. $\endgroup$ – lhf Feb 9 '12 at 13:29

$\zeta(2)$ is almost equal to $\sqrt e$ because if it weren't you'd be asking why it's almost equal to $\log_{10}44$. Honestly, interesting numbers are $\epsilon$-dense in the reals, where $\epsilon$ depends on what you find interesting, so it's guaranteed there will be interesting numbers close to each other, for no deeper reason at all.

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    $\begingroup$ Not all numerical coincidences are uninteresting. See en.wikipedia.org/wiki/Almost_integer for instance. Some are just coincidences, other have deeper reasons behind them. $\endgroup$ – lhf Feb 9 '12 at 13:51
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    $\begingroup$ You are correct, and my answer may have been unnecessarily harsh. $\endgroup$ – Gerry Myerson Feb 9 '12 at 22:50
  • $\begingroup$ One could cite "coincidences" related to Heegner numbers as "interesting". For ex.: $e^{\pi\sqrt{163}}$ $\endgroup$ – Klangen Sep 11 '17 at 13:14
  • $\begingroup$ @Pickle, that example is given, along with many more, in the link in the comment of lhf. $\endgroup$ – Gerry Myerson Sep 11 '17 at 13:17

Here is one way Lucky Larry might figure the limit probability that a number is squarefree: Larry already knows that $\zeta(2) \le 2 = 1+\sum_{1 \le n} \frac{1}{n \cdot (n+1)}$, which means no more than than half of all integers are squareful. Let $F_{n}:\{1,2,...,n\}\rightarrow\{1,2,...,2 \cdot n\}$ be a randomly chosen function whose range consists of all squareful integers between $1$ and $2 \cdot n$. So the condition of being squarefree is equivalent to not being in the range, and since the function is randomly chosen, the probability is $(1-\frac{1}{2 \cdot n})^n \rightarrow \frac{1}{\sqrt{e}}$. But of course the correct answer is $\frac{6}{\pi^2}$.

The problem with the above is that it ignores the constraint "whose range consists of all squareful numbers between $1$ and $2 \cdot n$" and the formula used only applies when the function is uniformly chosen. The almost-equality shows that this isn't always as big a mistake as it may seem.

  • $\begingroup$ I voted to reopen since I now have that privilege and I am still interested in other perspectives on this question (for example those explicated in the main comment thread). $\endgroup$ – Dan Brumleve May 14 '12 at 23:43

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