{n!}}\sum_{\ell=0}^n (-1)^\ell \ell!
{n \choose \ell} \ln^{n-\ell} a ~\mbox{Li}_{\ell+2}(-a)$$ $$=\sum_{\ell=0}^\infty (-1)^\ell \ell! \mbox{Li}_{\ell+2}(-a)\sum_{n=\ell}^\infty
{{(n+1)} \over {n!}}\sum_{\ell=0}^n {n \choose \ell}\ln^{n-\ell} a$$ $$=a \sum_{\ell=0}^\infty (-1)^\ell (1+\ln a +\ell)\mbox{Li}_{\ell+2}(-a)$$ $$=a\sum_{j=1}^\infty (-a)^j \left[{1 \over {(j+1)^2}}+{{\ln a} \over {j(j+1)}}
\right]$$ $$=-a+\ln a[-a+(1+a)\ln(a+1)]+\mbox{Li}_2(-a).  \eqno(70)$$ Combining terms completes the Corollary.

Akin to Eq. (24) we have from Eqs. (5) and (6) the weak upper asymptotic bounds $$|c_j| = O[(j-1)!], ~~~~~~|d_j(a)|=O[\Gamma(j,\ln a)], ~~j \to \infty.  \eqno(71)$$ These bounds miss cancellation, so are large over estimates.

Finally in closing this section we write a summation form for the constants $d_j(a)$. From Eq. (5) and the use of Lemma 1 we have $$d_j(a)=\sum_{k=0}^\infty \int_{k+a}^{k+a+1} {{\ln^{j-1} u} \over u} \left[\ln(k+a+1)-\ln(u+1) \right]du$$ $$={1 \over j}\sum_{k=0}^\infty \left[\ln^j(k+a+1)-\ln^j(k+a) \right]\ln(k+a+1)$$ $$-\sum_{k=0}^\infty\sum_{\ell=0}^{j-1}(-1)^\ell \ell!{{j-1} \choose \ell}\left[
\ln^{j-\ell-1}(k+a+1) \mbox{Li}_{\ell+2}(-k-a-1)-\ln^{j-\ell-1}(k+a) \mbox{Li}_{\ell+2}(-k-a)\right].  \eqno(72)$$

**Other expressions for $\gamma_0(a)$**

Recalling that $\gamma_0(a)=-\psi(a)$, we have [**Proposition 2**]{}. We have (a) $$\gamma_0(a)=-{1 \over a}+\ln(a+1)-\sum_{m=2}^\infty {{(-1)^m} \over m}
\left[\zeta(m,a)-{1 \over a^m}\right ],  \eqno(73)$$ (b) $$\sum_{m=2}^\infty{{(-1)^m} \over {m+1}}\zeta(m,a)=-a\ln a+\ln \Gamma(a+1)
-{1 \over 2} \psi(a)-{1 \over 2}\ln(2 \pi)+a,  \eqno(74)$$ and for $|z|<|a|$ (c) $$\sum_{m=2}^\infty{{z^m} \over m}\left[\zeta(m,a)-{1 \over a^m}\right]
={z \over a}+\ln\left(1-{z \over a}\right)-\ln \Gamma(a-z)+\ln\Gamma(a)
-z\psi(a).  \eqno(75)$$

Proof. For part (a), we form $$\sum_{m=2}^\infty {{(-1)^m} \over m}\left[\zeta(m,a)-{1 \over a^m}\right ]
=\sum_{m=2}^\infty {{(-1)^m} \over m} \sum_{k=1}^\infty {1 \over {(k+a)^m}}$$ 