$$=\sum_{k=1}^\infty \left[{1 \over {k+a}}-\ln(k+a+1)+\ln(k+a)\right]$$ $$=\ln(1+a)+\lim_{x \to \infty}\left[\sum_{k=1}^x {1 \over {k+a}}
-\ln([x]+a+1)\right] \eqno(76)$$ $$=\ln(1+a)-\psi(a+1)+\lim_{x \to \infty}\left[\psi([x]+a+1)-\ln([x]+a+1)
\right]$$ $$=\ln(1+a)-\psi(a+1)+\lim_{x \to \infty} O\left({1 \over {[x]+a+1}}\right)$$ $$=\ln(1+a)-\psi(a)-{1 \over a}.  \eqno(77)$$ In the last step we have used the functional equation of the digamma function.

An alternative method of proving part (a) is by applying the integral representation (35). Then we have $$\sum_{m=2}^\infty {{(-1)^m} \over m}\left[\zeta(m,a)-{1 \over a^m}\right ]
=\sum_{m=2}^\infty {{(-1)^m} \over m}\left[{1 \over {(m-1)!}}\int_0^\infty
{{t^{m-1} e^{-(a-1)t}} \over {e^t-1}}dt - {1 \over a^m}\right]$$ $$=-{1 \over a}+\ln\left(1 + {1 \over a}\right)-\sum_{m=2}^\infty {{(-1)^m} \over
{m!}} \int_0^\infty {{t^{m-1} e^{-(a-1)t}} \over {e^t-1}}dt$$ $$=-{1 \over a}+\ln\left(1 + {1 \over a}\right)-\int_0^\infty {{e^{-(a-1)t}} \over
{e^t-1}}{1 \over t}(e^{-t}+t-1)dt$$ $$=-{1 \over a}+\ln\left(1 + {1 \over a}\right)-\ln a + \psi(a)$$ $$=-{1 \over a}+\ln(a+1)+\psi(a).  \eqno(78)$$ The absolute convergence of the integral representation (35) justifies the interchange of summation and integration above.

For part (b), we have $$\sum_{m=2}^\infty {{(-1)^m} \over {m+1}}\zeta(m,a)=\sum_{m=2}^\infty {{(-1)^m} 
\over {(m+1)}} \sum_{k=0}^\infty {1 \over {(k+a)^m}}$$ $$=\sum_{k=0}^\infty\left\{(k+a)[\ln(k+a+1)-\ln(k+a)]+{1 \over 2}{1 \over {(k+a)}}
-1 \right\}$$ $$=\lim_{x \to \infty}\sum_{k=0}^x \left\{(k+a+1)\ln(k+a+1)-(k+a)\ln(k+a)
-\ln(k+a+1)+{1 \over 2}{1 \over {(k+a)}}-1 \right\} \eqno(79)$$ $$=\lim_{x \to \infty}\left\{([x]+a+1)\ln([x]+a+1)-a \ln a+\ln \Gamma(a+1)
-\ln\Gamma([x]+a+2) \right.$$ $$\left.+{1 \over 2}[\psi([x]+a+1)-\psi(a)]-([x]+1) \right \}.  \eqno(80)$$ We then apply Stirling’s formula for $\ln \Gamma(z)$ and part (b) follows.

For part (c), we proceed 