found from the special case of the Beta function $B(x,y)$ integral $$\int_0^1 (1-x)^2 x^{q-1}dx=B(3,q)={2 \over {q(q+1)(q+2)}}={1 \over q}
-{2 \over {q+1}}+{1 \over {q+2}},  \eqno(50)$$ by integrating over $q$. Similarly we may determine Eq. (43) by using the other integral of Eq. (46a).

Other forms of the sum of Eq. (44a) may be obtained from Eq. (46b) and the generating function $$g_j(z) \equiv \sum_{n=1}^\infty n^j (H_n-1)z^n, ~~~~~~~~~|z| < 1.  \eqno(51)$$ We may advance to $g_{j+1}$ from $g_j$ with the use of the operator $\theta
\equiv z (d/dz)$: $g_{j+1}(z) = \theta g_j(z)$. We have $g_1(z)=-z \ln(1-z)/(1-z)^2$ and obtain $$\sum_{k=1}^\infty {{\ln(k+1)} \over {k(k+1)}}=\sum_{n=1}^\infty n(H_n-1)\left[
{{\ln n} \over n}-2{{\ln(n+1)} \over {n+1}}+{{\ln(n+2)} \over {n+2}}\right]. \eqno(52)$$ One could pose the question whether all such expressions obtainable from the use of $g_j(z)$ are also obtainable through partial summation. Finally, we illustrate the application of a Binet formula for the digamma function [@grad] (p. 943), that we use as a representation of $\ln(k+1)$. We have $$\sum_{k=1}^\infty {{\ln(k+1)} \over {k(k+1)}}=\sum_{k=1}^\infty{{\psi(k+1)}
\over {k(k+1)}}+{1 \over 2}\sum_{k=1}^\infty{1 \over {k(k+1)^2}}+2\sum_{k=1}^
\infty \int_0^\infty {t \over {[t^2+(k+1)^2]}}{{dt} \over {(e^{2\pi t}-1)}}.
\eqno(53)$$ We evaluate the $\psi$ sum of this equation by inserting an integral representation for it and interchanging operations. For the sum on the right 