Example 4.19. Investigate the convergence of the series

$$
\sum_{n=1}^{\infty} \frac{p(p+1) \ldots(p+n-1)}{n!} \frac{1}{n^{q}}
$$
Solution. Apply Raabe's test:

$$
\begin{gathered}
\lim _{n \rightarrow \infty} n\left(\frac{a_{n}}{a_{n+1}-1}\right)= \\
=\lim _{n \rightarrow \infty} n\left(\frac{p(p+1) \ldots(p+n-1)}{n!} \frac{1}{n^{q}} \frac{(n+1)!(n+1)^{q}}{p(p+1) \ldots(p+n)}\right)= \\
=\lim _{n \rightarrow \infty} n\left(\frac{(n+1)^{q}}{n^{q}} \frac{n+1}{p+n}-1\right)= \\
=\lim _{n \rightarrow \infty} \frac{n}{n+p}\left(\left(1+\frac{1}{n}\right)^{q}(n+1)-p-n\right)= \\
=\lim _{n \rightarrow \infty} \frac{n}{n+p}\left(\left(1+\frac{q}{n}+o\left(\frac{1}{n}\right)\right)(n+1)-p-n\right)= \\
=\lim _{n \rightarrow \infty} \frac{n}{n+p}\left(1+q\right), \text{ i.e., when } q>p \text{ the series will be convergent, and when } 1+q-p<1, \text{ i.e., when } q<p \text{ the series will be divergent.}
$$