Using Gauss's test, examine the convergence of: sum
Using Gauss's test, examine the convergence of: $$ \sum_{n=1}^{\infty}\left[\frac{p(p+1) \cdots(p+n-1)}{q(q+1) \cdots(q+n-1)}\right]^\alpha, \qquad p>0,\; q>0. $$
Explanation of Method
Writing the term with Pochhammer symbols, $u_n=\left[\frac{(p)_n}{(q)_n}\right]^\alpha$, the ratio $u_n/u_{n+1}=\left[\frac{q+n}{p+n}\right]^\alpha$ expands (via the geometric-type series $\frac1{1+x}\approx1-x+x^2$ combined with the binomial expansion for the power $\alpha$) to give a clean leading coefficient $h=\alpha(q-p)$ in Gauss's test, generalizing Problem 87's hypergeometric threshold to an arbitrary power $\alpha$.
- Compute the ratio $u_n/u_{n+1}=\left[\frac{q+n}{p+n}\right]^\alpha$.
- Expand to leading order to extract $h=\alpha(q-p)$.
- Apply the convergence criterion in terms of $\alpha$, $p$, and $q$.
Step 1 · Compute the ratio
With $(p)_{n+1}=(p)_n(p+n)$ and likewise for $q$, $u_{n+1}=u_n\left[\frac{p+n}{q+n}\right]^\alpha$, so:
$$ \frac{u_n}{u_{n+1}} = \left[\frac{q+n}{p+n}\right]^\alpha. $$
Step 2 · Expand for large $n$
$\dfrac{q+n}{p+n}=\dfrac{1+q/n}{1+p/n}=1+\dfrac{q-p}n+O(1/n^2)$, and by the binomial expansion $(1+x)^\alpha=1+\alpha x+O(x^2)$:
$$ \frac{u_n}{u_{n+1}} = 1 + \frac{\alpha(q-p)}{n} + O(1/n^2). $$
Step 3 · Apply Gauss's test
Here $h=\alpha(q-p)$. So: converges if $\alpha(q-p)>1$; diverges if $\alpha(q-p)\le1$ (in particular, whenever $q\le p$, since then $h\le0<1$ automatically).
Summary Table
| Condition | Conclusion |
|---|---|
| $q>p$ and $\alpha>\dfrac1{q-p}$ | Converges |
| $q=p$ | Diverges ($h=0$) |
| $q<p$ | Diverges ($h<0$) |
| $q>p$, $\alpha\le\dfrac1{q-p}$ | Diverges |
Additional Notes
- Setting $\alpha=1$ recovers a special case of Problem 87's hypergeometric threshold, with the correspondence $\gamma-\alpha_{\text{hyp}}-\beta_{\text{hyp}}+1\leftrightarrow q-p$ under an appropriate identification — the underlying mechanism (Pochhammer-symbol ratios producing a clean linear Raabe coefficient) is identical.
- The requirement is purely on the product $\alpha(q-p)$, not on $\alpha$ or $q-p$ individually: a small gap $q-p$ can always be compensated by a sufficiently large power $\alpha$, and vice versa — a genuine trade-off between the two parameters.
- Since $(p)_n/(q)_n\sim C\,n^{p-q}$ for large $n$ (a standard Gamma-function asymptotic), $u_n\sim C^\alpha n^{\alpha(p-q)}$, and comparison with the $p$-series threshold $\alpha(q-p)>1$ confirms the Gauss's test result independently.
Result
The series converges if and only if $\alpha(q-p)>1$: