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Conjecture: \begin{align} \lim_{n\to \infty } \, \frac{\left(\prod _{k=1}^n \phi (k)\right){}^{1/n}}{n}\sim 0.2059\text{...} \end{align}

The numerical result from 100000 terms is:

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enter image description here

My questions are:

1) exists this limit ?

2) if yes, what is closed form of this constant ?

I am sure that this limit is NOT equal to 6/Pi^2 * exp(-1) = 0.2236438825...

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Yes and it equals $\frac{1}{e}\prod_p (1-1/p)^{1/p} $. – Lucia Feb 5 at 18:04
Nice result, thank you! – Vaclav Kotesovec Feb 5 at 18:42
up vote 10 down vote accepted

We may just write down $\varphi(k)=k\cdot \prod_{p|k} (1-1/p)$, $p$ runs over the primes which divide $k$, then $$ \frac{\left(\prod_{k\leqslant n}\varphi(k)\right)^{1/n}}{n}=\frac{\sqrt[n]{n!}}n\prod_{p} (1-1/p)^{\frac{[n/p]}{n}}. $$ First multiple tends to $1/e$ by weak version of Stirling's formula, and in the product we may replace each exponent $\frac{[n/p]}n$ to $1/p$ for $p\leqslant n$ with multiplicative error between 1 and $$ \prod_{p\leqslant n} (1-1/p)^{1/n}\geqslant \prod_{2\leqslant k\leqslant n}(1-1/k)^{1/n}=n^{-1/n}\rightarrow 1. $$ Hence the limit exists and equals $e^{-1}\prod_p (1-1/p)^{1/p}$ as Lucia and so-called friend Don have already said.

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Yes, the limit exists, and is equal to $\exp(-1 + \sum_{p} \frac{1}{p}\log(1-\frac{1}{p}))$ (as Lucia said), where $p$ runs over primes. Hint: Reduce the problem to computing $\lim_{N\to \infty} \frac{1}{N} \sum_{n \le N} \log \frac{\phi(n)}{n}$. The arithmetic function $\log\frac{\phi(N)}{N}$ is strongly additive and (under favorable conditions which hold here) the mean value of a strongly additive function $g$ is given by $\sum_{p} g(p)/p$.

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