# Integrating a Tricky Infinite Sum with a Rising Factorial

The sum that I need help integrating is as follows:
$$\int^{k}_{1}\frac{1}{n}+\frac{1}{n(n+k)}+\frac{1}{n(n+k)(n+2k)}+\frac{1}{n(n+k)(n+2k)(n+3k)}+\ …$$
I was unable to find information on how to do a general integration of a rising factorial to obtain a closed form. If anyone could point me in the right direction I would greatly appreciate it.

#### Solutions Collecting From Web of "Integrating a Tricky Infinite Sum with a Rising Factorial"

By partial fraction decomposition we have
$$\frac{1}{n(n+k)}=\frac{1}{k}\left(\frac{1}{n}-\frac{1}{n+k}\right),\qquad \frac{1}{n(n+k)(n+2k)} = \frac{1}{2k^2}\left(\frac{1}{n}-\frac{2}{n+k}+\frac{1}{n+2k}\right)$$
and in general
$$\frac{1}{n(n+k)\cdots(n+mk)}=\frac{1}{m! k^m}\sum_{j=0}^{m}\frac{(-1)^j \binom{m}{j}}{n+jk}\tag{1}$$
It follows that the contribute given by the pole at $n=0$ equals
$$\int_{1}^{k}\frac{dn}{n}\sum_{m\geq 0}\frac{1}{m!k^m}\,dn = e^{1/k}\log k\tag{2}$$
the contribute given by the pole at $n=-k$ equals
$$\int_{1}^{k}\frac{dn}{n+k}\sum_{m\geq 1}\frac{(-1)m}{m!k^m}=e^{1/k}\frac{1}{k}\log\left(\frac{1+k}{2k}\right)\tag{3}$$
the contribute given by the pole at $n=-2k$ equals
$$\int_{1}^{k}\frac{dn}{n+2k}\sum_{m\geq 2}\frac{\binom{m}{2}}{m!k^m}=e^{1/k}\frac{1}{2k^2}\log\left(\frac{3k}{2k+1}\right)\tag{4}$$
and so on.