Using Cauchy's root test, examine the convergence of: sum
Using Cauchy's root test, examine the convergence of: $$ \sum_{n=1}^{\infty}\left(\frac{x}{n}\right)^n, \qquad x>0. $$
Explanation of Method
Here $\sqrt[n]{b_n}=\dfrac xn$ simplifies almost instantly, since the $n$-th root cancels the $n$-th power exactly, leaving a bare ratio. As $n\to\infty$ this ratio tends to $0$ for any fixed $x$, so the series converges regardless of how large $x$ is chosen.
- Apply the root test to $b_n=(x/n)^n$.
- Evaluate $\rho=\lim_{n\to\infty}x/n$.
- Confirm convergence for all fixed $x>0$, with numerical checks.
Step 1 · Apply the root test
With $b_n=\left(\dfrac xn\right)^n$:
$$ \sqrt[n]{b_n} = \sqrt[n]{\left(\frac xn\right)^n} = \frac xn. $$
Step 2 · Evaluate the limit
Since $x$ is a fixed positive constant:
$$ \rho = \lim_{n\to\infty}\frac xn = 0. $$
Step 3 · Apply the convergence criterion
Since $\rho=0<1$, by Cauchy's root test the series converges for every $x>0$. Intuitively, once $n>x$, the ratio $x/n<1$, and raising a number below $1$ to an ever-larger power drives it to zero rapidly — e.g. for $x=10,\,n=20$: $b_{20}=(1/2)^{20}\approx9.5\times10^{-7}$; for $x=100,\,n=200$: $b_{200}=(1/2)^{200}\approx6.2\times10^{-61}$.
Summary Table
| Quantity | Value |
|---|---|
| $\sqrt[n]{b_n}$ | $x/n$ |
| $\rho=\lim\sqrt[n]{b_n}$ | $0$, for every $x>0$ |
| Conclusion | Converges for all $x>0$ |
Additional Notes
- This is one of the cleanest possible root-test computations: the $n$-th root and the $n$-th power cancel exactly, with no need for logarithms or L'Hôpital's rule, unlike Problem 71's $n/(\ln n)^n$.
- The convergence is uniform in the sense that no matter how large $x$ is fixed, the same argument $\rho=\lim x/n=0$ applies — only the number of "small" initial terms before the rapid decay sets in grows with $x$.
- Problem 73 generalizes this exact setup by replacing the fixed denominator $n$ with a general sequence $a_n\to a$, turning this problem's automatic convergence into a threshold condition on $x$ versus $a$.
Result
The series converges for every $x>0$: