Show $\Sigma_{n\leq X} 1/\phi(n) \sim \log(X)*\Sigma_{k=1}^\infty \mu(k)^2/(k*\phi(k))$ 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.

Found an answer.  $\sum_{t\leq X/k} 1/t = \log(X/k) + \gamma + O(1/(X/k))$  When you multiply it out, the sum can be split into four terms:
$$ \sum_{n\leq X} 1/\phi(n) = \log(X)*\sum_{k\leq X} \mu(k)^2/(k*\phi(k)) - \sum_{k\leq X} \mu(k)^2*\log(k)/(k*\phi(k)) $$
$$ + \sum_{k\leq X} \gamma*\mu(k)^2/(k*\phi(k)) + \sum_{k\leq X} \mu(k)^2/(k*\phi(k)*O(k/X)) $$
I had also previously shown that $\sum_{k\leq X} \mu(k)^2/(k*\phi(k))$ and $\sum_{k\leq X} \mu(k)^2*\log(k)/(k*\phi(k))$ converge, so as X approaches infinity, the last three terms of the sum become constants, leaving only the first term.  Thus we can conclude that
$$ \lim_{X\to \infty} \frac{\sum_{n\leq X} 1/\phi(n)}{\sum_{k\leq X} \mu(k)^2/(k*\phi(k))} = 1 $$
