Sign up ×
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.

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.

share|cite|improve this question

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

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

They're really not that close. –  Qiaochu Yuan Feb 9 '12 at 7:26
I checked this also, and I found numerically that there are infinitely many real numbers between $\zeta(2)$ and $\sqrt{e}$. –  yohBS Feb 9 '12 at 7:29
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 $u\to0$. –  Did Feb 9 '12 at 7:36
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 $a=2/3\lt1$). –  Did Feb 9 '12 at 9:27
1.64872... and 1.64493... are not that close at all. –  lhf Feb 9 '12 at 13:29

2 Answers 2

$\zeta(2)$ 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.

share|cite|improve this answer
Not all numerical coincidences are uninteresting. See for instance. Some are just coincidences, other have deeper reasons behind them. –  lhf Feb 9 '12 at 13:51
You are correct, and my answer may have been unnecessarily harsh. –  Gerry Myerson Feb 9 '12 at 22:50
up vote 5 down vote accepted

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.

share|cite|improve this answer
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). –  Dan Brumleve May 14 '12 at 23:43

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