Using D'Alembert's ratio test, determine the convergence
Using D'Alembert's ratio test, determine the convergence of: $$ \sum_{n=1}^{\infty} \frac{n^p}{n!}. $$
Explanation of Method
The ratio simplifies to $\left(1+\frac1n\right)^{p-1}\cdot\frac1n$; since $\left(1+\frac1n\right)^{p-1}\to1^{p-1}=1$ for any fixed real $p$, the entire ratio is governed by the factor $\frac1n\to0$. No matter how large $p$ is chosen, factorial growth in the denominator eventually overwhelms the polynomial numerator, so the series converges universally.
- Compute the ratio $a_{n+1}/a_n$ for general $p$.
- Evaluate the limit, showing it equals $0$ regardless of $p$.
- Apply the ratio test to conclude universal convergence.
Step 1 · Compute the ratio
With $a_n=\dfrac{n^p}{n!}$:
$$ \frac{a_{n+1}}{a_n} = \frac{(n+1)^p}{(n+1)!}\cdot\frac{n!}{n^p} = \frac{(n+1)^p}{(n+1)\cdot n^p} = \frac{(n+1)^{p-1}}{n^{p-1}}\cdot\frac1n = \left(1+\frac1n\right)^{p-1}\cdot\frac1n. $$
Step 2 · Evaluate the limit
For any fixed real $p$: $\left(1+\frac1n\right)^{p-1}\to1^{p-1}=1$, so:
$$ L = \lim_{n\to\infty} \left(1+\frac1n\right)^{p-1}\cdot\frac1n = 1\cdot0 = 0. $$
Step 3 · Apply the ratio test
Since $L=0<1$ for every real $p$, by D'Alembert's ratio test $\sum_{n=1}^{\infty}\dfrac{n^p}{n!}$ converges for all $p\in\mathbb R$.
Summary Table
| Quantity | Value |
|---|---|
| Ratio $a_{n+1}/a_n$ | $\left(1+\dfrac1n\right)^{p-1}\cdot\dfrac1n$ |
| $L=\lim a_{n+1}/a_n$ | $0$, for every real $p$ |
| Conclusion | Converges for all $p\in\mathbb R$ |
Additional Notes
- Special cases recover familiar results: $p=0$ gives $\sum1/n!=e-1$ (Problem 58); $p=1$ gives $\sum n/n!=e$; $p=2$ gives $\sum n^2/n!=2e$ — all consequences of differentiating or manipulating the exponential series.
- Even negative $p$ works: e.g. $p=-1$ gives $\sum\frac1{n\cdot n!}$, which converges by the same argument, since the ratio test's conclusion never actually depended on the sign of $p$.
- This universal convergence — regardless of how large $p$ is — makes $n!$ the strongest possible "convergence enforcer" among common denominator growth rates, stronger even than any fixed exponential $a^n$ combined with any polynomial power.
Result
The series converges for every real $p$: