Using Raabe's test, examine the convergence of: sum

Using Raabe's test, examine the convergence of: $$ \sum_{n=1}^{\infty} \frac{\alpha(\alpha+1) \cdots(\alpha+n-1)\,\beta(\beta+1) \cdots(\beta+n-1)}{n!\,\gamma(\gamma+1) \cdots(\gamma+n-1)}. $$

Step 1 · Set up $u_n$ and $u_{n+1}$

With $u_n=\dfrac{(\alpha)_n(\beta)_n}{n!(\gamma)_n}$, using $(\alpha)_{n+1}=(\alpha)_n(\alpha+n)$ and likewise for $\beta,\gamma,n!$:

$$ u_{n+1} = \frac{(\alpha)_n(\alpha+n)(\beta)_n(\beta+n)}{n!(n+1)(\gamma)_n(\gamma+n)}. $$

Step 2 · Compute the ratio

$$ \frac{u_n}{u_{n+1}} = \frac{(n+1)(\gamma+n)}{(\alpha+n)(\beta+n)}. $$

Step 3 · Apply Raabe's test

Expanding both products: $(n+1)(\gamma+n)=n^2+n(1+\gamma)+\gamma$ and $(\alpha+n)(\beta+n)=n^2+n(\alpha+\beta)+\alpha\beta$, so:

$$ \frac{u_n}{u_{n+1}}-1 = \frac{n(\gamma-\alpha-\beta+1)+(\gamma-\alpha\beta)}{(\alpha+n)(\beta+n)}, \qquad r = \lim_{n\to\infty} n\left(\frac{u_n}{u_{n+1}}-1\right) = \gamma-\alpha-\beta+1. $$

So: converges if $\gamma>\alpha+\beta$; diverges if $\gamma<\alpha+\beta$; at $\gamma=\alpha+\beta$ ($r=1$), a finer test (Gauss's test) shows the series still diverges.

Summary Table

Additional Notes

• This is exactly Gauss's classical criterion for the hypergeometric series ${}_2F_1(\alpha,\beta;\gamma;x)$ evaluated at the boundary point $x=1$ of its unit radius of convergence, one of the most historically significant applications of Raabe-type tests.
• Problem 90's series $\sum\frac{(a)_n}{n!}\cdot\frac1{n^a}$ can be seen as a degenerate special case of this hypergeometric structure (formally $\beta\to\infty$ absorbed into the extra factor $n^{-a}$), which is why both problems land on Raabe limit $r=1$ or a similarly delicate boundary.
• The symmetry between $\alpha$ and $\beta$ in the convergence criterion ($\gamma>\alpha+\beta$, not favoring either) reflects the underlying symmetry of the hypergeometric function itself: ${}_2F_1(\alpha,\beta;\gamma;x)={}_2F_1(\beta,\alpha;\gamma;x)$.