On the mean value of a multiplicative function: Prove that $\sum\limits_{n\leq x} \frac{n}{\phi(n)} =O(x) $ There is a second part of the problem posted in Proving $ \frac{\sigma(n)}{n} < \frac{n}{\varphi(n)} < \frac{\pi^{2}}{6} \frac{\sigma(n)}{n}$, from Apostol's book, but I can't figure it out. It asks the reader to prove that if $x \geq 2$ then $\sum\limits_{n\leq x} \frac{n}{\phi(n)} = O(x)$, where the right hand side of the equation uses the Big-O notation. I've tried using the prior result, but I always end up with $O(x^2)$ . Any insights?
 A: @ Jane
The following might be what Apostol had in mind:
Considering conjugate divisors we may write $$\frac{\sigma(n)}{n}=\sum_{c|n}\frac{1}{c}$$
and hence $$\sum_{n \leq x}\frac{\sigma(n)}{n}=\sum_{c \leq x}\frac{1}{c}\left[\frac{x}{c}\right]\leq x \zeta(2)$$ so that $$\sum_{n\leq x} \frac{n}{\phi(n)} \leq \zeta(2)^2 x$$ It should be noted that the mean of $\frac{\phi(n)}{n}$ always stays close to some constant when considered over any reasonably large set of numbers.
A: We can actually prove the slightly stronger $$\sum_{n\leq x}\frac{n}{\phi (n)}=\frac{315}{2\pi^4}\zeta(3)x +O\left(\log x\right).$$ 
Idea: Notice that $\frac{n}{\phi(n)}$ is very close in some sense to $1$.  So if we write it as $1*f$ for some $f$, and this $f$ should be very small.  The following lemma, which is one of the first was you might see to sum under the hyperbola, then tells us exactly how to deal with this situation:

If $(1*f)(n)=\sum_{d|n}f(d)$ denotes Dirichlet convolution, then $$\sum_{n\leq x} (1*f)(n)=\sum_{n\leq x}\sum_{d|n}f(d)=\sum_{d\leq x} f(d)\left[\frac{x}{d}\right]$$ and using $[z]=z+O(1)$ this is $$=x\sum_{d\leq x} \frac{f(d)}{d}+O\left(\sum_{d\leq x} |f(d)|\right).$$

Since we need only look at prime powers, we need $f$ to satisfy $$(1*f)(p^k)=1+f(p)+\cdots+f(p^k)=\frac{p^k}{\phi(p^k)}=\frac{p}{p-1}.$$  This is constant with respect to $k$, so we take $f(p^k)=0$ for $k\geq 2$, and then $f(p)=\frac{1}{p-1}$.  This can be rewritten for all $n$ as $f(n)=\frac{\mu(n)^2}{\phi(n)}$.  Hence, using the above we see that $$\sum_{n\leq x}\frac{n}{\phi (n)}=x\sum_{d\leq x} \frac{\mu(d)^2}{d\phi(d)}+O\left(\sum_{d\leq x}\frac{\mu(d)^2}{\phi(d)}\right).$$
The big-O term term is bounded by $\log (x)$, and the infinite series $\sum_{d=1}^\infty \frac{\mu(d)^2}{d\phi(d)}$ converges absolutely.  Playing around with the Euler product, we can actually find that this equals $\frac{\zeta(2)\zeta(3)}{\zeta(6)}=\frac{315}{2\pi^4}\zeta(3)$. Extending the series up to $x$ all the way to infinitely creates an error of the form $\frac{\log \log x}{x}$ which is negligible.   Hence we conclude 
$$\sum_{n\leq x}\frac{n}{\phi (n)}=\frac{315}{2\pi^4}\zeta(3)x +O\left(\log x\right).$$ 
Remark: You are looking at the mean value of a multiplicative function.  For functions "close" to one, there is a simple generalization of the above, but much can be said about general bounded functions, even those which are not close to $1$.  Of course this is more difficult, as the mean value of $\mu(n)$ is related to the prime number theorem, but there is much interesting work.  Specifically Halasz's theorem tells us exactly when the mean value can be  large.  Granville and Soundararajan have developed this, and used it as an alternative starting point for proving theorems in analytic number theory.
Edit: To evaluate the series $$\sum_{d=1}^\infty \frac{\mu(d)^2}{d\phi(d)},$$ we may first note that since the summand is a multiplicative funciton, and since everything converges nicely, we can write this as an Euler product $$\sum_{d=1}^{\infty}\frac{\mu(d)^{2}}{d\phi(d)}=\prod_{p}\left(1+\frac{\mu(p)^{2}}{p\phi(p)}+\frac{\mu(p^{2})^{2}}{p^{2}\phi(p^{2})}+\cdots\right).$$ This equals 
$$\prod_{p}\left(1+\frac{1}{p\left(p-1\right)}\right)=\prod_{p}\left(\frac{p^{2}-p+1}{p\left(p-1\right)}\right).$$ Recalling the factorization $p^{3}+1=\left(p^{2}-p+1\right)\left(p+1\right),$ multiply to and bottom by $p+1$ to get a $p^3+1$ term on top.  Since we hope to rewrite things in terms of the zeta function, multiply top and bottom by $p^{3}-1$ as well. We then get $$\prod_{p}\left(\frac{p^{3}+1}{p(p^{2}-1)}\cdot\frac{p^{3}-1}{p^{3}-1}\right)=\prod_{p}\left(1-\frac{1}{p^{6}}\right)\prod_{p}\left(1-\frac{1}{p^{3}}\right)^{-1}\prod_{p}\left(1-\frac{1}{p^{2}}\right)^{-1},$$ which equals $\frac{\zeta(2)\zeta(3)}{\zeta(6)}.$
See Also:


*

*An answer which works out the exact details for multiplicative functions close to $n$.

*This technique used on a more complicated sum involving four convolutions of the mobius function

*An answer on the error term that arises when taking the average of the Totient function.

*Averages of certain arithmetic functions.
A: if you prove that $\phi(d) \gg \frac{d}{\log d}$ then the convergence of the series is easy to see. The stronger $\phi(d) \gg \frac{d}{\log \log d}$ can be proved using the Chebyshev's estimates $\frac{x}{\log x}\ll \pi(x) \ll \frac{x}{\log x}$. A one line proof can be given from these estimates for the first inequality as follows : Every prime below $d$ is coprime to $d$ or divides $d$ and hence $\pi(d)-\omega(d) \leq \phi(d)$ from which it follows that  $\frac{d}{\log d} +O(\log d) \ll \phi(d)$ which is what you want. The estimate $\omega(d) \ll \log d $ can be obtained by taking logarithms in $2^{\omega(d)}\leq d.$ 
