Earlier today I thought I proved that the following series diverged:


as a result of a misapplication of the prime number theorem. I mistook $\phi(n)$ for $\pi(n)$ in the statement. Is this salvageable? I've been trying to find lower bounds for $\phi(n)$ that I may pass to, but the literature on this is a bit dense for me.


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    $\begingroup$ I think you want an upper bound, to show the terms are less than those of some convergent series. $\endgroup$ – Chappers Apr 20 '17 at 22:13
  • $\begingroup$ Thanks, I edited the statement. The misapplication would show divergence. $\endgroup$ – Prototank Apr 20 '17 at 22:14

It diverges. In fact, just taking the terms corresponding to primes gives $$\sum_{p \text{ prime}} \frac{\phi(p)}{p^2} = \sum_{p \text{ prime}} \frac{p-1}{p^2} > \sum_{p \text{ prime}} \frac{1}{2p}$$ and the sum of the reciprocals of the primes is known to diverge (Wikipedia link).

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    $\begingroup$ I was going this way, but you got there first. (+1). $\endgroup$ – Chappers Apr 20 '17 at 22:16
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    $\begingroup$ So if the Prime Number Theorem is given, in the form of $\lim_{k\to\infty} \frac{p_k}{k\log(k)}=1$, then does your idea, combined with limit comparison test, show that the series diverges? (I love your answer by the way, it is short and sweet. I merely have an audience in mind and would like to use something familiar, or at least famous enough) $\endgroup$ – Prototank Apr 20 '17 at 22:57
  • $\begingroup$ Sure: given $\lim_{k \to \infty} \frac{p_k}{k \log k} = 1$, we know that $\lim_{k \to \infty} \frac{1/(2p_k)}{1/(2k \log k)} = 1$, so $\sum_{p \text{ prime}} \frac{1}{2p} = \sum_{k=1}^\infty \frac{1}{2p_n}$ diverges iff $\sum_{k=1}^\infty \frac{1}{2k \log k}$ diverges (which is most straightforwardly done by integral test). $\endgroup$ – Misha Lavrov Apr 20 '17 at 23:00
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    $\begingroup$ The divergence of the sum of reciprocals of the primes goes back to Euler, and in particular precedes the Prime Number Theorem. One way is that $$ \sum_{n=1}^N \frac{1}{n} \le \prod_{p_k \le N} \frac{1}{1-1/p_k}$$ and an infinite product $\prod_k \frac{1}{1-a_k}$ with $0 < a_k < 1$ diverges iff $\sum_k a_k$ diverges. $\endgroup$ – Robert Israel Apr 20 '17 at 23:06

Wikipedia says:

$$\sum_{n=1}^\infty\frac{\phi(n)}{n^s} = {\zeta(s-1)\over\zeta(s)}$$

So for $s = 2$, we get $\zeta(1)/\zeta(2)$ which diverges.


We have $\varphi(n)\leq n$, hence the Dirichlet series $\sum_{n\geq 1}\frac{\varphi(n)}{n^s}$ is convergent for any $s>2$.
Since $\varphi$ is a multiplicative function, by Euler's product $$ \sum_{n\geq 1}\frac{\varphi(n)}{n^s} = \prod_{p}\left(1+\frac{\varphi(p)}{p^s}+\frac{\varphi(p^2)}{p^{2s}}+\ldots\right)=\prod_{p}\frac{1-p^{-s}}{1-p^{1-s}}=\frac{\zeta(s-1)}{\zeta(s)} $$ for any $s>2$, where the RHS behaves like $\frac{6}{\pi^2(s-2)}$ as $s\to 2^+$.
It follows that the original series is divergent: $$ \sum_{n=1}^{N}\frac{\varphi(n)}{n^2}\approx \frac{6}{\pi^2}\log N.$$


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