Using D'Alembert's ratio test, determine the convergence
Using D'Alembert's ratio test, determine the convergence of: $$ \sum_{n=1}^{\infty} \frac{a^n}{n!}, \qquad a>0. $$
Explanation of Method
The ratio $\frac{a_{n+1}}{a_n}=\frac a{n+1}\to0$ for every fixed $a>0$, since the growing factorial in the denominator eventually outpaces any fixed exponential base $a$ in the numerator, no matter how large $a$ is. This universal convergence identifies the series with (a shifted version of) the Taylor expansion of $e^a$.
- Compute the ratio $a_{n+1}/a_n$ and its limit for a fixed $a$.
- Apply the ratio test, noting the conclusion holds for every $a>0$.
- Identify the sum as $e^a-1$.
Step 1 · Compute the ratio
With $a_n=\dfrac{a^n}{n!}$:
$$ \frac{a_{n+1}}{a_n} = \frac{a^{n+1}}{(n+1)!}\cdot\frac{n!}{a^n} = \frac{a}{n+1} \longrightarrow 0. $$
Step 2 · Apply the ratio test
Since $L=0<1$ regardless of the fixed value of $a>0$, by D'Alembert's ratio test $\sum_{n=1}^{\infty}\dfrac{a^n}{n!}$ converges for every $a>0$.
Step 3 · Identify the sum
Since $e^a=\sum_{n=0}^{\infty}\frac{a^n}{n!}$, subtracting the $n=0$ term ($=1$):
$$ \sum_{n=1}^{\infty}\frac{a^n}{n!} = e^a - 1. $$
Summary Table
| Quantity | Value |
|---|---|
| Ratio $a_{n+1}/a_n$ | $\dfrac a{n+1}$ |
| $L=\lim a_{n+1}/a_n$ | $0$, for every $a>0$ |
| Sum | $e^a-1$ |
Additional Notes
- Even for extreme values like $a=10$, the series still converges (to $e^{10}-1\approx22{,}025$) — the point at which the ratio $\frac{a}{n+1}$ drops below $1$ just gets pushed to a larger $n$, but it always eventually does, since $n+1\to\infty$.
- This result directly contrasts with Problem 59's $\sum2^n/n$: replacing the linear denominator $n$ with a factorial $n!$ is the single change that flips divergence into universal convergence, illustrating how much faster factorial growth is than linear growth.
- Problem 63's series $\sum n!(x/n)^n$ is a much more delicate variant of this same idea, where the roles of exponential and factorial growth are pitted against each other differently, producing a finite critical threshold ($x=e$) rather than universal convergence.
Result
The series converges for every $a>0$, with sum: