\eqno(85)$$ Then we have the expressions $$M_j(k)={1 \over {(j+2)}}{1 \over {(k+a)^2}} ~_2F_1\left(1,j+2;j+3;-{1 \over
{k+a}}\right) \eqno(86)$$ $$=(-1)^j (k+a)^jM_0(k)-(-1)^j(k+a)^j\sum_{m=2}^{j+1}{{(-1)^m}
\over m} {1 \over {(k+a)^m}}, \eqno(87)$$ where $_2F_1$ is the Gauss hypergeometric function and $$M_0(k)={1 \over {k+a}}-\ln\left(1+{1 \over {k+a}}\right).  \eqno(88)$$

For the proof of Eq. (86), let $(a)_n=\Gamma(a+n)/\Gamma(a)$ be the Pochhammer symbol. By noting that $(j+2)_m/(j+3)_m=(j+2)/(j+m+2)$ we have $$M_j(k)=\sum_{m=0}^\infty {{(-1)^m} \over {(m+j+2)}}{1 \over {(k+a)^{m+2}}}
={1 \over {(j+2)}}{1 \over {(k+a)^2}}\sum_{m=0}^\infty {{(1)_m} \over {m!}}
{{(j+2)_m} \over {(j+3)_m}} \left({{-1} \over {k+a}}\right)^m.  \eqno(89)$$ By the series definition of the function $_2F_1$, Eq. (86) holds. In particular, $M_0(k)$ is a known special case.

For Eq. (87), we observe by shifting the summation index that $$M_{j+1}(k)=-(k+a)M_j(k) + {1 \over {(j+2)}}{1 \over {(k+a)}}.  \eqno(90)$$ In addition to this recursion relation we have $$M_j(k)=(-1)^j(k+a)^j \sum_{m=j+2}^\infty {{(-1)^m} \over m}{1 \over {(k+a)^m}},
\eqno(91)$$ and Eq. (87) follows.

For the proof of Proposition 3, we write $$\sum_{m=2}^\infty {{(-1)^m} \over {m+j}}\zeta(m,a)=\sum_{k=0}^\infty M_j(k),
\eqno(92)$$ and then apply Lemma 4.

The manner in which the sums of terms rational in $k+a$ in Proposition 3 may be evaluated in terms of polygamma functions $\psi^{(j)}$ is given by [**Lemma 5**]{}. We have (a) $$\sum_{k=0}^x (k+a)^{j-m}={{(-1)^{m-j}} \over {(m-j-1)!}}\left[\psi^{(m-j-1)}
(a)-\psi^{(m-j-1)}([x]+a+1)\right],  \eqno(93)$$ and (b) the asymptotic value of the second term on 