Convergence of $\sum\limits_{n=1}^{\infty}{{n!n^{-p}}\over{q(q+1)...(q+n)}}$ Convergence of $\sum\limits_{n=1}^{\infty}{{n!n^{-p}}\over{q(q+1)...(q+n)}}$
In this post can you explain how first step is done?
I understand that we need to use Gauss test here but  can't understand how to get this
${{n+1}\over{q+n+1}} \left({{n}\over{n+1}}\right)^p
= \left( 1 - \frac qn  + O(1/n^2)\right)
\left( 1 - \frac pn  + O(1/n^2)\right)
 $
 A: 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

*

*If $\lim_{n\to\infty}n^pa_n<\infty$ and $p>1$, then $S$ is convergent.

*If $\lim_{n\to\infty}n^pa_n\not= 0$ and $p\le 1$, then $S$ is divergent.

