I would like to show that as X approaches infinity,
$\sum_{n\leq X} \frac{1}{\phi(n)} \sim \log(X)\cdot\sum_{k=1}^{\infty} \frac{\mu(k)^2}{k\cdot\phi(k)}.$
I have already proven
$\sum_{n\leq X} \frac{1}{\phi(n)} = \sum_{k\leq X} \left[\frac{\mu(k)^2}{k\cdot\phi(k)}\cdot\sum_{t\leq X/k} \frac1t\right].$
I recognize that $\Sigma_{t\leq X/k} 1/k$ is the $X/k$th harmonic number and that harmonic numbers can be approximated by $\ln(X/k)$.
Wolfram Alpha says that $\sum_{k=1}^{\infty} \frac{\mu(k)^2}{k\cdot\phi(k)} = \frac{315\zeta(3)}{2\pi^4}.$
I have a feeling an epsilon-delta proof might be appropriate here, but I'm not sure where to proceed.