I am reading Tenenbaum's "Introduction to Analytic Number Theory", and I am confused about the utility of the second part of one of the core theorems in the Selberg-Delange method, as it applies to counting problems. Here is the full theorem stated for clarity:
Theorem 3: Let $a_z(n)$ be an arithmetic function depending on a complex paramater $z$ and with a power series expansion in the disc $|z|\leq A$ $$a_z(n)=\sum_{k=0}^{\infty}c_k(n)z^k$$ Let $N$ be a non-negative integer. Suppose that there exist $N+1$ functions $h_j(z)$ (0\leq j\leq N), holomorphic for $|z|\leq A$, and a quantity $R_N(x)$, independent of $z$, such that for $x\geq 3$ and $|z|\leq A$ $$\sum_{n\leq x}a_z(n)=x(\log(x))^{z-1}\left[\sum_{j=0}^N\frac{zh_j(z)}{(\log(x))^j}+O_A(R_N(x))\right]$$ Then uniformly for $x\geq 3$, $1\leq k \leq A\log_2(x)$ we have $$\sum_{n\leq x} c_k(n)=\left[\sum_{j=0}^N\frac{Q_{j,k}(\log_2(x))}{(\log(x))^j}+O_A\left(\frac{(\log_2(x))^k}{k!}R_N(x)\right)\right]$$ with $$Q_{j,k}(X):=\sum_{m+l=k-1}\frac{1}{m!l!}h_j^{(m)}(0)X^l$$ If, in addition, we suppose that $|h_0''(z)|\leq B$ for $|z|\leq A$, then, uniformly for $x\geq 3$, $1\leq k\leq A\log_2(x)$, we have $$\sum_{n\leq x}c_k(n)=\frac{x}{\log(x)}\frac{(\log_2(x))^{k-1}}{(k-1)!}\left[h_0\left(\frac{k-1}{\log_2(x)}+O_A\left(\frac{B(k-1)}{(\log_2(x))^2}+\frac{\log_2(x)}{k}R_0(x)\right)\right)\right]$$
The form required for the hypothesis of the theorem is very specific, and from what I understand its only real use is in conjunction with another theorem that will give an asymptotic for $\sum_{n<x}a_z(n)$ in the desired form. The error term for this theorem, however, will be at best
$$R_N(x)=e^{-c_1\sqrt{\log(x)}}+\left(\frac{c_2N+1}{\log(x)}\right)^{N+1}$$
and thus
$$R_N(x)=O(\log^{-1}(x))$$
which would give an error term of $O\left(\frac{1}{k}\right)$ in the second part of the theorem, which is hardly useful since one would want to be guaranteed that the error in the approximation goes to zero as $x\to\infty$. Am I missing something? When would this second part of the theorem be useful?