The following thread at math.stackexchange.com proposes a constant term for the asymptotic expansion of $\sum_{1\le k \le n} \frac{\varphi(k)}{k^2}.$
I am getting a different term and I would like to know which is right.
My calculation goes like this: first recall the classic identity $ \sum_{d\mid n} \varphi(d) = n,$ which gives the Dirichlet convolution $ \varphi \star 1 = n \quad \text{and hence} \quad \sum_{n\ge 1} \frac{\varphi(n)}{n^s} = \frac{\zeta(s-1)}{\zeta(s)}.$ To evaluate the sum use the Mellin-Perron type integral $\int_{2-i\infty}^{2+i\infty} A(s) n^s \frac{ds}{s} = -\frac{1}{2}\frac{\varphi(n)}{n^2} + \sum_{k=1}^n \frac{\varphi(k)}{k^2} \quad \text{where} \quad A(s) = \frac{\zeta(s+1)}{\zeta(s+2)}.$ and shift to the left to pick up the residue at $s=0$, getting $ \frac{6}{\pi^2} \log n + \left(\frac{6\gamma}{\pi^2} - \frac{36}{\pi^4} \zeta'(2)\right) + O\left( \frac{\log n}{n} \right).$
Addendum. In view of Eric Naslunds excellent comment below (the second one to point out the missing details) maybe we can ask whether anyone is able to supply those missing bounds on the rest of the contour for the Mellin-Perron integral, thereby turning this question into a useful reference. Here is a MSE challenge of the same type.