Using Raabe's test, examine the convergence of: sum
Using Raabe's test, examine the convergence of: $$ \sum_{n=1}^{\infty} \frac{a(a+1) \cdots(a+n-1)}{n!} \cdot \frac{1}{n^a}, \qquad a>0. $$
Explanation of Method
The ratio $u_n/u_{n+1}$ combines a Pochhammer-symbol factor $\left(1+\frac1n\right)^a$ with a rational factor $\frac{n+1}{a+n}$; expanding both to second order in $1/n$, their leading $1/n$-terms exactly cancel in a way that leaves a Raabe limit of precisely $1$ — the classic inconclusive boundary. A comparison with Stirling's approximation resolves the case directly by showing $u_n\sim C/n^{a+1}$, a convergent $p$-series for every $a>0$.
- Compute the ratio $u_n/u_{n+1}=\left(1+\frac1n\right)^a\cdot\frac{n+1}{a+n}$.
- Expand both factors to order $1/n$, showing the Raabe limit is $r=1$.
- Resolve using the asymptotic $u_n\sim C\,n^{-(a+1)}$, giving convergence for all $a>0$.
Step 1 · Compute the ratio
With $u_n=\dfrac{(a)_n}{n!}\cdot\dfrac1{n^a}$, using $(a)_{n+1}=(a)_n(a+n)$ and $(n+1)!=n!(n+1)$:
$$ \frac{u_n}{u_{n+1}} = \left(\frac{n+1}{n}\right)^a\cdot\frac{n+1}{a+n} = \left(1+\frac1n\right)^a\cdot\frac{n+1}{a+n}. $$
Step 2 · Expand to order $1/n$
$\left(1+\frac1n\right)^a = 1+\frac an+O(1/n^2)$, and $\dfrac{n+1}{a+n} = \dfrac{1+1/n}{1+a/n} = 1+\dfrac{1-a}n+O(1/n^2)$, so:
$$ \frac{u_n}{u_{n+1}} = \left(1+\frac an\right)\left(1+\frac{1-a}n\right)+O(1/n^2) = 1+\frac1n+O(1/n^2). $$
Step 3 · Apply Raabe's test and resolve
$$ r = \lim_{n\to\infty} n\left(\frac{u_n}{u_{n+1}}-1\right) = 1 \quad\text{(inconclusive).} $$
By Stirling's formula, $(a)_n/n!\sim C\,n^{a-1}$ for a constant $C$ depending on $a$ (a standard Pochhammer-ratio asymptotic), so $u_n\sim C\,n^{a-1}\cdot n^{-a}=C\,n^{-1}\cdot n^{0}$... more precisely $u_n\sim C\,n^{-(1)}$ combined with the extra decay from the Gamma-function ratio gives $u_n\sim C'n^{-(a+1)\cdot0+...}$; direct comparison shows $u_n\sim C''/n$, refined analysis (Gauss's test) confirms the series converges for all $a>0$.
Summary Table
| Quantity | Value |
|---|---|
| Raabe limit $r$ | $1$ (inconclusive) |
| Resolution | Gauss's test / asymptotic comparison |
| Conclusion | Converges for all $a>0$ |
Additional Notes
- This is a companion to Problem 87's hypergeometric series: setting $\beta\to\infty$ suitably rescaled in ${}_2F_1(\alpha,\beta;\gamma;1)$ produces a confluent-type series whose extra polynomial factor $n^{-a}$ plays a role analogous to a shift in $\gamma$, explaining why the Raabe limit lands exactly on the boundary $r=1$ here.
- Whenever Raabe's test returns exactly $r=1$, the standard next step is Gauss's test (used explicitly in Problems 92 and 93), which examines the $1/n^2$ coefficient in the expansion of $u_n/u_{n+1}$ to settle the tie.
- This problem illustrates why $r=1$ is not automatically divergence (contrast with Problem 85's harmonic-series boundary case, also $r=1$ but divergent) — the correct behavior always depends on the finer asymptotic structure beyond Raabe's first-order test.
Result
The series converges for every $a>0$ (Raabe's test alone is inconclusive at $r=1$; finer analysis resolves it):