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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.

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    They're really not that close.2012-02-09
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    I checked this also, and I found numerically that there are infinitely many real numbers between $\zeta(2)$ and $\sqrt{e}$.2012-02-09
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    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 $u\to0$.2012-02-09
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    Is there any pattern to 8/3, 31/9, 141/23 or are you just picking rational numbers to suit your desires? Also, it's not that $\zeta(x)$ approaches $\exp(2^{-x})$, it's that both approach $1$.2012-02-09
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    @anon, just picking. Only the $8/3$ is really surprising to me but maybe there is a pattern there; I included $141/23$ to show that the best denominator isn't always $3^n$. Anyway I'm interested in any serious arguments that these almost-equalities are or are not surprising.2012-02-09
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    A serious argument about a subjective reaction? At any rate, you can find rationals arbitrarily close to $\zeta^{-1}(e^{1/n})$ (this is hardly particular to $\zeta$ and $\exp$) and there is nothing special about your choice versus other choices of rationals, so in such an absence of anything unusual the burden of proof is not on the skeptics' side. In order to explain a coincidence, you need there to be a co-incidence in the first place, and you haven't pointed to what two things are co-inciding here.2012-02-09
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    @anon, I think the observations are surprising in the same way as $e^\pi - \pi \approx 20$ and $\{e^{\sqrt{163}}\} \approx 0$. I'll retag as soft-question.2012-02-09
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    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 $a=2/3\lt1$).2012-02-09
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    1.64872... and 1.64493... are not that close at all.2012-02-09
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    ...because $\small | 36 e-\pi^4 | <0.5 $ ?...2012-02-09
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    My vote to close the question was swayed in particular by the argument that many numbers will be close to each other, and by the fact that the discussion seems to be not so productive. However, I think that Didier Piau's first comment above makes the best out of the question and I found it enlightening.2012-02-11
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    I'd still like to know why $2/3$ is the best constant in that formula for small values of $n$. I don't understand all the stingy interpretations, oh well, at least I came up with a partial answer myself before the closure.2012-02-11
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    @CarlMummert Thanks.2012-02-12
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    Solving $\zeta(n+s_n)=\mathrm e^{1/2^n}$ yields the expansion $s_n=\frac1{\log2}a^n +\frac1{2\log2}b^n+o\left(b^n\right)$ with $a=\frac23$ and $b=\frac12$. Considering that $\frac1{\log2}$ is within $4\%$ of $\frac32$, this might explain the seemingly good fit of $s_n$ with $\left(\frac23\right)^{n-1}=\frac32a^{n}$.2012-02-12

2 Answers 2

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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.

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    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).2012-05-14
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$\zeta(2)$ 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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    Not all numerical coincidences are uninteresting. See http://en.wikipedia.org/wiki/Almost_integer for instance. Some are just coincidences, other have deeper reasons behind them.2012-02-09
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    You are correct, and my answer may have been unnecessarily harsh.2012-02-09
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    One could cite "coincidences" related to Heegner numbers as "interesting". For ex.: $e^{\pi\sqrt{163}}$2017-09-11
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    @Pickle, that example is given, along with many more, in the link in the comment of lhf.2017-09-11