the right side of Eq. (93) is given by $$-{{(-1)^{m-j}} \over {(m-j-1)!}}\psi^{(m-j-1)}([x]+a+1)
={1 \over {(m-j-1)}}{1 \over {([x]+a+1)^{m-j-1}}}+O\left({1 \over {([x]+a+1)^{m-j}}}
\right),$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x \to \infty. \eqno(94)$$

For the proof of part (a) of this Lemma, we note that $$\sum_{k=0}^x (k+a)^{j-m}=\zeta(m-j,a)-\zeta(m-j,[x]+a+1).  \eqno(95)$$ We then use the relation $$\zeta(n+1,x)={{(-1)^{n+1}} \over {n!}} \psi^{(n)}(x), ~~~~~~~n \geq 1.
\eqno(96)$$ For part (b) we apply the known asymptotic form of the polygamma functions [@nbs].

Remarks. The asymptotic evaluation of terms of the form $(k+a)^j[\ln(k+a+1)-\ln(k+a)]$ in Proposition 3 may be performed by way of Euler-Maclaurin summation. So at $j=1$ one way to write the result is in terms of the Glaisher-Kinkelin constant. Similarly, at $j=2$ and $j=3$ the results may be written in terms of constants $B$ and $C$ that have previously appeared in the literature (e.g., [@sri], pp. 36 and 37).

Examination of Eq. (95) shows that $j$ there need not be an integer. We may therefore differentiate it with respect to $j$ and find $$\sum_{k=0}^x (k+a)^{j-m}\ln(k+a)=\zeta'(m-j,[x]+a+1)-\zeta'(m-j,a),
\eqno(97)$$ where $'$ denotes differentiation with respect to the first argument. Also for use in connection with Proposition 3 we have by the integral representation (35) and Euler-Maclaurin summation $\zeta(s,a+x)=x^{1-s}/(s-1)+O(x^{-s})$ as $x \to \infty$, giving $$\zeta'(s,a+x)={x^{1-s} \over {s-1}}\ln x - {x^{1-s} 