Using Raabe's test, examine the convergence of: sum
Using Raabe's test, examine the convergence of: $$ \sum_{n=1}^{\infty} \frac{n!}{(a+1)(a+2) \cdots(a+n)}. $$
Explanation of Method
Raabe's test is needed here because D'Alembert's ratio test itself gives $\frac{u_{n+1}}{u_n}\to1$ (inconclusive), so a finer measurement of how fast the ratio approaches $1$ is required. Computing $n\left(\frac{u_n}{u_{n+1}}-1\right)$ turns out to have the remarkably clean limit $a$, so the whole problem reduces to comparing the parameter $a$ against Raabe's threshold of $1$.
- State Raabe's test.
- Compute the ratio $u_n/u_{n+1}$ using the rising-factorial structure.
- Evaluate the Raabe limit $r=a$ and resolve the boundary case $a=1$ directly.
Step 1 · State Raabe's test
For $u_n>0$, let $r=\lim_{n\to\infty} n\left(\dfrac{u_n}{u_{n+1}}-1\right)$: if $r>1$ the series converges, if $r<1$ it diverges, if $r=1$ the test is inconclusive.
Step 2 · Compute the ratio $u_n/u_{n+1}$
With $u_n=\dfrac{n!}{(a+1)(a+2)\cdots(a+n)}$ and $u_{n+1}=\dfrac{(n+1)!}{(a+1)\cdots(a+n)(a+n+1)}$:
$$ \frac{u_n}{u_{n+1}} = \frac{n!\,(a+n+1)}{(n+1)!} = \frac{a+n+1}{n+1}. $$
Step 3 · Evaluate the Raabe limit and resolve $a=1$
$$ \frac{u_n}{u_{n+1}}-1 = \frac{a}{n+1} \quad\Longrightarrow\quad r = \lim_{n\to\infty} n\cdot\frac{a}{n+1} = \lim_{n\to\infty}\frac{a}{1+\frac1n} = a. $$
So $a>1\Rightarrow r>1$, converges; $a<1\Rightarrow r<1$, diverges. At $a=1$: $u_n=\dfrac{n!}{2\cdot3\cdots(n+1)}=\dfrac{n!}{(n+1)!}=\dfrac1{n+1}$, so the series is $\sum\frac1{n+1}$ — the harmonic series shifted by one term — which diverges.
Summary Table
| Value of $a$ | $r$ | Conclusion |
|---|---|---|
| $a>1$ | $a>1$ | Converges |
| $a=1$ | $1$ | Diverges (harmonic series) |
| $a<1$ | $a<1$ | Diverges |
Additional Notes
- The special case $a=2$ gives $u_n=\dfrac{2}{(n+1)(n+2)}\sim\dfrac2{n^2}$, matching the $p$-series convergence threshold ($p=2>1$) as an independent sanity check on the Raabe result $r=2>1$.
- The special case $a=0$ gives $u_n=\dfrac{n!}{n!}=1$ for every $n$, so $\sum1$ trivially diverges — consistent with $r=0<1$ from the formula, even though $a=0$ falls slightly outside the series' most natural domain $a>-1$ (needed so no factor $a+k$ vanishes).
- D'Alembert's ratio test alone would only establish $u_{n+1}/u_n\to1$ here (inconclusive at every $a$), making this problem a clean illustration of why Raabe's test exists: it extracts the coefficient of the $1/n$ correction term that the plain ratio test discards.
Result
The series converges if and only if $a>1$: