on the right side of Eq. (17) gives $$j\sum_{\ell=0}^{j-1} (-1)^{\ell-1}\ell! {{j-1} \choose \ell}\ln^{j-\ell-1} x ~
\mbox{Li}_{\ell+2}(-x)$$ $$\sim j\sum_{\ell=0}^{j-1} (-1)^{\ell-1}\left[-{{\ln^{j+1} x}
\over {(\ell+1)(\ell+2)}}+2\sum_{k=1}^{[\ell+2]/2} {{\ln^{\ell+2-2k} x} \over
{(\ell+2-2k)!}}\mbox{Li}_{2k}(-1)\right]$$ $$={j \over {j+1}}\ln^{j+1} x$$ $$+2j\ln^{j+1}x\sum_{\ell=0}^{j-1} (-1)^{\ell-1}
{{j-1} \choose \ell} \sum_{k=1}^{[\ell+2]/2} {{\ell!} \over {(\ell+2-2k)!}}
\ln^{-2k} x \mbox{Li}_{2k}(-1).  \eqno(21)$$ For a given $k$, the sum over $\ell$ on the right side of Eq. (21) gives zero as $$\sum_{\ell=0}^{j-1} (-1)^{\ell-1}{{j-1} \choose \ell}{{\ell!} \over
{(\ell+2-2k)!}}=-{1 \over {(j-2k+1)}}{1 \over {\Gamma(2-2k)}}$$ $$=-{1 \over {(j-2k+1})\pi}\Gamma(2k-1)\sin \pi(2k-1).  \eqno(22)$$

We now have from Eqs. (11), (12), and (21) that $$\sum_{m=2}^\infty {{(-1)^m} \over m}(-1)^j \zeta^{(j)}(m)
=\lim_{x \to \infty}\left\{\sum_{k=2}^x {{\ln^j k} \over k}-\ln^{j+1} x
+O\left({{\ln^j x} \over x}\right)\right.$$ $$\left.+{j \over {j+1}}\ln^{j+1} x +j c_j(x)
+(-1)^j j!(1-2^{-j})\zeta(j+1)\right\}$$ $$=\lim_{x \to \infty} \left(\sum_{k=2}^x {{\ln^j k} \over k}-{{\ln^{j+1} x}
\over {j+1}}\right)+jc_j+(-1)^j j!(1-2^{-j})\zeta(j+1)$$ $$=\gamma_j +jc_j+(-1)^j j!(1-2^{-j})\zeta(j+1).  \eqno(23)$$ In the last step we used relation (2). For the next to last step, since we have $$\ln\left({{[t]+1} \over {t+1}}\right )=O\left({{\{t\}} \over t}\right),
\eqno(24)$$ where $\{t\}$ denotes the fractional part of $t$, the constants $c_j$ are certainly well defined and $\lim_{x \to \infty} c_j^{(x)} = c_j$. We have demonstrated part (a).

For part (b) we proceed similarly to part (a). We first have $$\sum_{m=2}^\infty {{(-1)^m} \over m}(-1)^j \zeta^{(j)}(m,a)
=\lim_{x \to \infty}\left \{\sum_{k=0}^x {{\ln^j 