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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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    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

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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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    @Pickle, that example is given, along with many more, in the link in the comment of lhf.2017-09-11