I'm trying to provide a full-fledged proof of the estimate

$$\Psi(x,x^{1/u}) = x \, \rho(u)+ O\left(\frac{x}{\log x}\right) \qquad (*)$$

via the sly inductive approach commonly attributed to A. Granville. (For any $U\geq 0$ the formula holds uniformly for $u \in [0,U]$ and all $x\geq 2$; the $\rho$ in it denotes the Dickman function.) As many of you know, Prof. Granville begins his *proof* by noticing that $(*)$ holds for every $u \in(0,2]$ and every $x\geq 2$ and by assuming afterwards that it also holds for every $u \in (0,N]$ and every $x\geq 2$. Then, resorting to the Buchstab identity

$$\Psi(x,y) = 1 + \sum_{p \leq y} \Psi \left(\frac{x}{p},p\right),$$

he obtains that

$$\Psi(x, x^{1/u}) = \Psi(x,x^{1/N})- \sum_{x^{1/u}<p\leq x^{1/N}}\Psi\left(\frac{x}{p},p\right). \qquad (**)$$

At this point, the idea is to invoke the induction hypothesis and rewrite every term in the sum on the right-hand side of $(**)$. Given that

$$p = \left(\frac{x}{p}\right)^{\log p / \log(x/p)},$$

it follows that if $p>x^{1/u}$ and $u \in (N,N+1]$, then

$$\frac{\log(x/p)}{\log p} = \frac{\log x}{\log p}-1 < u-1 \leq N$$

and whence

\begin{eqnarray*} \Psi(x,x^{1/u}) &=& x \, \rho(N) + O\left(\frac{x}{\log x}\right) - \sum_{x^{1/u}<p\leq x^{1/N}}\left[\frac{x}{p} \, \rho\left(\frac{\log x}{\log p}-1\right) + O\left(\frac{x/p}{\log(x/p)}\right)\right]\\ &=& x \, \rho(N) + O\left(\frac{x}{\log x}\right) + O\left(\frac{x}{\log x}\sum_{x^{1/u}<p\leq x^{1/N}} \frac{1}{p}\right) - x\sum_{x^{1/u}<p\leq x^{1/N}}\frac{1}{p} \, \rho\left(\frac{\log x}{\log p}-1\right)\\ &=& x \, \rho(N) + O\left(\frac{x}{\log x}\right) - x\sum_{x^{1/u}<p\leq x^{1/N}}\frac{1}{p} \, \rho\left(\frac{\log x}{\log p}-1\right).\\ \end{eqnarray*}

Notice that in going from the next-to-last line to the last one, we resorted to the famed second theorem of F. Mertens. Having said all this, my question is the following one:

What's in your experience the most clear-cut way to deal with the remaining sum in the last line of the math environment preceding the previous sentences?

By recurring to Merten's second theorem (once again) and the Abel summation formula, I am able to recover the expected main term but I don't see a way to *sense* in advance if proceeding thus one is to obtain a *suitable* error term (I eventually obtain it; yet, one usually prefers to find out whether he/she is on track before actually getting there, right?).

I would greatly appreciate any insights, suggestions, or comments you may want to leave for me below.