Earlier today I thought I proved that the following series diverged:
$$\sum_{n=2}^\infty\frac{\phi(n)}{n^2}$$
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.
Thanks
On
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.
On
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:
$$ \sum_{n=1}^{N}\frac{\varphi(n)}{n^2}\approx \frac{6}{\pi^2}\log N.$$
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).