Convergence of $\sum\limits_{n=1}^{\infty}{{n!n^{-p}}\over{q(q+1)...(q+n)}}$

Solution 1:

1)There is an alternative and easier way to check for convergence as we will see soon. About the operations from the post, if $p\in\mathbb{N}$, look at this:

\begin{eqnarray} \frac{n+1}{q+n+1}&=&1-\frac{q}{n+q+1}\\ &=& 1 -\frac{q}{n}+\frac{q}{n}-\frac{q}{n+q+1}\\ &=& 1-\frac{q}{n} + \frac{q^2+q}{n(n+q+1)}\\ &=& 1 - \frac{q}{n}+O(1/n^2) \end{eqnarray} since $\frac{n^2(q^2+q)}{n(n+q+1)}\to q^2+q$ when $n\to\infty$.
For the other factor using binomial theorem we got \begin{eqnarray} \Bigg(\frac{n}{n+1}\Bigg)^p &=& \Bigg(1-\frac{1}{n+1}\Bigg)^p\\ &=& 1 - \frac{p}{n+1} + \displaystyle \sum_{k=2}^p \binom{p}{k}\Bigg(-\frac{1}{n+1}\Bigg)^k\\ &=& 1-\frac{p}{n}+\frac{p}{n}- \frac{p}{n+1} + \displaystyle \sum_{k=2}^p \binom{p}{k}\Bigg(-\frac{1}{n+1}\Bigg)^k\\ &=& 1 - \frac{p}{n} + O(1/n^2) \end{eqnarray} since $\frac{pn^2}{n(n+1)}+n^2\sum_{k=2}^p \binom{p}{k}\Big(-\frac{1}{n+1}\Big)^k\to p+\binom{p}{2}$ as $n\to\infty$. Case when $p\in\mathbb{R_+}$ is similar.

2)The alternative way for checking convergence is using the Pringsheim's Criteria that states the following:
If $S=\sum_{n\ge 1}a_n$ is a series with non-negative terms, we have