Using D'Alembert's ratio test, determine the convergence
Using D'Alembert's ratio test, determine the convergence of: $$ \sum_{n=1}^{\infty} n!\left(\frac{x}{n}\right)^n, \qquad x>0. $$
Explanation of Method
The ratio here reduces to $x\cdot\left(\frac n{n+1}\right)^n$, and since $\left(\frac n{n+1}\right)^n\to\frac1e$, the limit is exactly $\frac xe$. This produces a genuine critical threshold at $x=e$, unlike Problem 60's universal convergence — a competition between the growing factorial and the growing power $x^n/n^n$ that resolves in favor of convergence only when $x<e$.
- Identify the general term and compute the ratio $a_{n+1}/a_n$.
- Evaluate the limit using $\left(\frac n{n+1}\right)^n\to\frac1e$.
- Analyze the boundary case $x=e$ separately using Stirling's approximation.
Step 1 · Identify the term and compute the ratio
Write $a_n = n!\left(\dfrac xn\right)^n = \dfrac{n!\,x^n}{n^n}$. Then:
$$ \frac{a_{n+1}}{a_n} = \frac{(n+1)!\,x^{n+1}/(n+1)^{n+1}}{n!\,x^n/n^n} = \frac{(n+1)\,x\,n^n}{(n+1)^{n+1}} = x\left(\frac n{n+1}\right)^n. $$
Step 2 · Evaluate the limit
Using $\left(\frac n{n+1}\right)^n=\dfrac1{(1+1/n)^n}\to\dfrac1e$:
$$ L = x\cdot\frac1e = \frac xe. $$
By the ratio test: converges if $\frac xe<1$ (i.e. $x<e$); diverges if $\frac xe>1$ (i.e. $x>e$); inconclusive at $x=e$.
Step 3 · Analyze the boundary case $x=e$
At $x=e$: $a_n = n!\left(\dfrac en\right)^n = \dfrac{n!\,e^n}{n^n}$. By Stirling's approximation $n!\sim\sqrt{2\pi n}\left(\dfrac ne\right)^n$:
$$ a_n \sim \sqrt{2\pi n}\left(\frac ne\right)^n\cdot\frac{e^n}{n^n} = \sqrt{2\pi n}. $$
Since $\sum\sqrt n$ diverges ($p=\frac12<1$), the series diverges at $x=e$.
Summary Table
| Value of $x$ | Convergence |
|---|---|
| $0<x<e$ | Converges |
| $x=e$ | Diverges (terms $\sim\sqrt{2\pi n}$) |
| $x>e$ | Diverges |
Additional Notes
- The constant $e$ arises here for the same reason it appears in Problem 61 ($\sum n!/n^n$): setting $x=1$ in this problem recovers exactly Problem 61's series, and indeed $1<e$ confirms the earlier convergence result as a special case.
- This is a genuinely delicate boundary case — unlike most $p$-series thresholds, where the boundary can go either way depending on a secondary exponent, here the boundary $x=e$ always diverges, because the leading correction from Stirling's formula ($\sqrt{2\pi n}$) always fails to be summable.
- The critical value $x=e$ is the same constant that appears throughout factorial asymptotics (Problems 16, 19, 49, 50, 61), reflecting that $n!\sim\sqrt{2\pi n}(n/e)^n$ is the universal source of these $e$-thresholds.
Result
The convergence of the series depends on $x$ relative to $e$: