Using Raabe's test, examine the convergence of: sum
Using Raabe's test, examine the convergence of: $$ \sum_{n=1}^{\infty} \frac{\sqrt{n!}}{(a+\sqrt{1})(a+\sqrt{2}) \cdots(a+\sqrt{n})}, \qquad a>0. $$
Explanation of Method
The ratio $u_n/u_{n+1}$ simplifies remarkably to $\dfrac{a+\sqrt{n+1}}{\sqrt{n+1}}=1+\dfrac{a}{\sqrt{n+1}}$ — a correction term of order $n^{-1/2}$ rather than the usual $n^{-1}$. Multiplying by $n$ in Raabe's formula therefore produces $n/\sqrt n=\sqrt n\to\infty$, so the Raabe limit is infinite and convergence holds regardless of $a$.
- Compute $u_{n+1}$ using $\sqrt{(n+1)!}=\sqrt{n+1}\cdot\sqrt{n!}$.
- Simplify the ratio $u_n/u_{n+1}=\dfrac{a+\sqrt{n+1}}{\sqrt{n+1}}$.
- Evaluate the Raabe limit, showing $r=\infty>1$.
Step 1 · Set up $u_n$ and $u_{n+1}$
With $u_n=\dfrac{\sqrt{n!}}{\prod_{k=1}^n(a+\sqrt k)}$ and $\sqrt{(n+1)!}=\sqrt{n+1}\cdot\sqrt{n!}$:
$$ u_{n+1} = \frac{\sqrt{n+1}\cdot\sqrt{n!}}{(a+\sqrt{n+1})\prod_{k=1}^n(a+\sqrt k)}. $$
Step 2 · Compute the ratio
$$ \frac{u_n}{u_{n+1}} = \frac{a+\sqrt{n+1}}{\sqrt{n+1}} = 1 + \frac{a}{\sqrt{n+1}}. $$
Step 3 · Apply Raabe's test
$$ r = \lim_{n\to\infty} n\cdot\frac{a}{\sqrt{n+1}} = \lim_{n\to\infty}\frac{a\sqrt n}{\sqrt{1+1/n}} = \infty. $$
Since $r=\infty>1$, by Raabe's test the series converges for every $a>0$.
Summary Table
| Quantity | Value |
|---|---|
| $u_n/u_{n+1}-1$ | $\dfrac{a}{\sqrt{n+1}}$ |
| Raabe limit $r$ | $\infty$ |
| Conclusion | Converges for all $a>0$ |
Additional Notes
- An infinite Raabe limit is a strong signal that the series converges "faster than any $p$-series" — indeed, since $u_n/u_{n+1}-1\sim a/\sqrt n$, the terms decay roughly like a rapidly accelerating geometric-type product rather than a mere power of $n$.
- The denominator's factors $a+\sqrt k$ each exceed $\sqrt k$, so the denominator grows faster than $\prod\sqrt k=\sqrt{n!}$, the same order as the numerator — the extra additive constant $a$ in every factor is precisely what tips the balance decisively toward convergence.
- Compare with Problem 85's $\sum\frac{n!}{(a+1)\cdots(a+n)}$, where the correction term was $O(1/n)$ and the Raabe limit was the finite value $a$ (converging only for $a>1$): here the square roots inside each factor slow the growth of the denominator's index just enough to produce an unboundedly large correction term instead, guaranteeing convergence unconditionally.
Result
The series converges for every $a>0$: