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I was playing around with the series $f(N)=\frac{1}{N}\sum_{k=1}^{N}\frac{\phi(k)}{k}$ and I found with Wolfram that $f(10,000)=0.607938$, which I noticed was very close to $\frac{6}{\pi^2}$.

I am led to make the following

Conjecture: $\lim_{N \to +\infty}\frac{1}{N}\sum_{k=1}^{N}\frac{\phi(k)}{k}=\frac{6}{\pi^2}$

Well, is it true?

Note that its obvious that the sum is bounded above by $1$ (since $\phi(k)/k<1$), so it definitely doesn't diverge to infinity. Its also almost always decreasing. So it most likely converges.

math_lover
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2 Answers2

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The key formula is $$\frac{\phi(n)}{n}=\sum_{d\mid n} \frac{\mu(d)}{d}$$

From this, you get that:

$$\begin{align}f(N)=&\sum_{k=1}^{N}\frac{\phi(k)}{k}\\&=\sum_{k=1}^{N}\sum_{d\mid k} \frac{\mu(d)}{d}\\ &= \sum_{d=1}^{N}\frac{\mu(d)}{d}\left\lfloor\frac N d\right\rfloor\end{align}$$

Then $\left|\left\lfloor\frac N d\right\rfloor - \frac{N}d\right|<1$ so $$\left|\frac{f(N)}N-\sum_{d=1}^{N}\frac{\mu(d)}{d^2}\right|<\frac{1}{N}\sum_{d=1}^{N}\frac1{d^2}< \frac{\zeta(2)}{N}$$

But we know that $$\sum_{d=1}^{\infty}\frac{\mu(d)}{d^2}=\frac{1}{\zeta(2)}=\frac{6}{\pi^2}$$

Thomas Andrews
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Here's a start:

It is well known that (a,b) = 1 for random a,b with probability $6/\pi^2.$ Then $\phi(k) \sim k \cdot \frac{6}{\pi^2}$ for large k, and because the density of primes is zero, by the definition of $\phi(x),$ so $$\frac{1}{n}\sum_{n\geq k\geq 1}\frac{\phi(k)}{k}\sim \frac{1}{n}\sum_{n\geq k\geq 1} \frac{6}{\pi^2}=\frac{6}{\pi^2}.$$ However this is not very rigorous.

Teoc
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    Yes but it is from this result that you know the density of $(a,b)=1$ is $6/\pi^2$ – reuns Dec 24 '16 at 02:18
  • @user1952009 : is it not from the Euler product formula and zeta(2)? – math_lover Dec 24 '16 at 02:27
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    Of course, it is only sort-of true that $\phi(k)/k\to \frac{6}{\pi^2}$. It is not actually true - for example, when $p_i$ are the primes, $\frac{\phi(p_i)}{p_i}\to 1$. Also $\phi(2^n)/2^n = \frac{1}{2}$ for all $n$. So it's just an idea, really – Thomas Andrews Dec 24 '16 at 03:06