side of Eq. (53) we apply the partial fraction decomposition for the hyperbolic cotangent function [@grad] (p. 36). We obtain $$\sum_{k=1}^\infty {{\ln(k+1)} \over {k(k+1)}}=1-\gamma+{\pi^2 \over {12}}
+2\int_0^\infty \left[{\pi \over {2t}}\coth \pi t-{1 \over {2t^2}}-{1 \over 
{t^2+1}}\right]{{t dt} \over {(e^{2\pi t}-1)}} \eqno(54a)$$ $$={3 \over 2}-2\gamma+{\pi^2 \over {12}}+\int_0^\infty\left(\pi \coth \pi t
-{1 \over t}\right){{dt} \over {(e^{2\pi t}-1)}}.  \eqno(54b)$$ For the last step we have used [@grad] (p. 328).

\(iii) The alternating sum on the right side of Eq. (9), $$\sum_{m=3}^\infty {{(-1)^m} \over m}\left({{m-2} \over {m-1}}\right)^n
\simeq \sum_{m=3}^\infty {{(-1)^m} \over m} e^{-n/(m-1)}, \eqno(55)$$ is exponentially decreasing with $n$. The last integral term on the right side of Eq. (9) is over estimated by $n-1$.

**A Corollary and further remarks**

The method of Proposition 1(b) also has relevance to Dirichlet $L$ functions, as these functions may be written as a linear combination of Hurwitz zeta functions. For instance, for $\chi$ a character modulo $m$ and Re $s > 1$ we have $$L(s,\chi) = \sum_{k=1}^\infty {{\chi(k)} \over k^s} ={1 \over m^s}\sum_{k=1}^m 
\chi(k) \zeta\left(s,{k \over m}\right).  \eqno(56)$$ Conversely, this relation may be inverted, yielding the Hurwitz zeta function $\zeta(s,a)$ at rational values of $a$ in terms of a linear combination of $L$ functions. When $\chi$ is 