as in the second proof of part (a). We have $$\sum_{m=2}^\infty {{z^m} \over m}\left[\zeta(m,a)-{1 \over a^m}\right ]
=\sum_{m=2}^\infty {{z^m} \over m}\left[{1 \over {(m-1)!}}\int_0^\infty
{{t^{m-1} e^{-(a-1)t}} \over {e^t-1}}dt - {1 \over a^m}\right]$$ $$={z \over a}+\ln\left(1 - {z \over a}\right)-\int_0^\infty {{e^{-(a-1)t}}
\over {(e^t-1)}}{1 \over t}(e^{zt}-zt-1)dt.  \eqno(81)$$ We use [@grad] (p. 939) $$\ln \Gamma(a-z)-\ln\Gamma(a)=\int_0^\infty \left[{{e^{-(a-z)t}-e^{-at}} \over
{1-e^{-t}}}-z e^{-t}\right]{{dt} \over t}, \eqno(82)$$ along with an integral representation of $\gamma$ plus the digamma function [@grad] (p. 943) and of the Euler constant [@grad] (p. 946) in the form $$\gamma=\int_0^\infty \left[{e^{-t} \over {1-e^{-t}}}-{e^{-t} \over t}\right]
dt.  \eqno(83)$$ Proposition 2 is finished.

Remark. Part (a) of this Proposition gives the $a \neq 1$ extension of Ref. [@verma]. Part (b) gives the $a \neq 1$ extension of Refs. [@singh] and [@suryan].

**Generalizations of sum of Hurwitz zeta function values**

Here we generalize the sums of the previous section, showing [**Proposition 3**]{}. For integers $j \geq 1$ we have $$\sum_{m=2}^\infty {{(-1)^m} \over {m+j}}\zeta(m,a)=(-1)^j \lim_{x \to \infty}
\sum_{k=0}^x \left\{(k+a)^{j-1}-(k+a)^j \ln\left(1+{1 \over {k+a}}\right)\right.$$ $$\left.-(k+a)^j \sum_{m=2}^{j+1} {{(-1)^m} \over m} {1 \over {(k+a)^m}} \right \}.
\eqno(84)$$

The proof of this Proposition depends upon [**Lemma 4**]{}. For integers $j \geq 0$ and $k \in C$ let $$M_j(k) \equiv \sum_{m=2}^\infty {{(-1)^m} \over {(m+j)}}{1 \over {(k+a)^m}}.
