Estimating a sum involving Euler's $\phi$ function

Question

For some $B > 2$ I want to estimate the following sum

$$\sum_{n=2}^{B}\frac{\phi (n)}{n^3} $$

As $B \not \to \infty$ I can't use the asymptotic estimate $\phi (n) \approx n$ and substitute into the sum, so I am looking for an alternative.

Answer 1

By Euler's product for Dirichlet's series $$ \sum_{n\geq 1}\frac{\varphi(n)}{n^{2s}}=\prod_{p\in\mathcal{P}}\left(1+\frac{\varphi(p)}{p^{2s}}+\frac{\varphi(p^{2})}{p^{4s}}+\ldots\right)=\prod_{p\in\mathcal{P}}\frac{1-p^{-2s}}{1-p^{1-2s}}=\frac{\zeta(2s-1)}{\zeta(2s)} $$ is a meromorphic function with a simple pole at $s=1$ with residue $\frac{6}{\pi^2}$. It follows that $$ \sum_{n=2}^{B}\frac{\varphi(n)}{n^2} \approx \frac{6}{\pi^2}\log(B) $$ for large values of $B$. The same can be proved by summation by parts, since the average order of $\varphi(n)$ is well-known. In a similar way

$$ \sum_{n\geq 1}\frac{\varphi(n)}{n^3} = \frac{\zeta(2)}{\zeta(3)} $$ and $$ \sum_{n=2}^{B}\frac{\varphi(n)}{n^3} = \frac{\zeta(2)}{\zeta(3)}-1+O\left(\frac{1}{B}\right).$$