Problem 73
Medium Convergence Tests

Using Cauchy's root test, examine the convergence of: sum

Using Cauchy's root test, examine the convergence of: $$ \sum_{n=1}^{\infty}\left(\frac{x}{a_n}\right)^n, $$ where $\lim_{n\to\infty} a_n = a$, $x>0$, and $a_n>0$.

Explanation of Method

This generalizes Problem 72 (where $a_n=n\to\infty$ forced automatic convergence) by letting $a_n$ tend to a finite limit $a$. The root test simplifies exactly to $\sqrt[n]{b_n}=x/a_n\to x/a$, so the outcome now depends on comparing $x$ to $a$, with a genuine threshold at $x=a$ and a separate degenerate case when $a=0$.

  1. Apply the root test to get $\sqrt[n]{b_n}=x/a_n$.
  2. Evaluate $\rho=x/a$ for $a>0$, and handle $a=0$ separately.
  3. Analyze the borderline case $x=a$.

Step 1 · Apply the root test

With $b_n=\left(\dfrac x{a_n}\right)^n$:

$$ \sqrt[n]{b_n} = \frac{x}{a_n}. $$

Step 2 · Evaluate the limit

Case $a>0$: since $a_n\to a$, $\rho=\lim_{n\to\infty}\dfrac x{a_n}=\dfrac xa$.

Case $a=0$: then $\rho=\lim_{n\to\infty}\dfrac x{a_n}=+\infty>1$, so the series diverges for every $x>0$.

Step 3 · Apply the criterion and analyze $x=a$

For $a>0$: converges if $x<a$ ($\rho<1$); diverges if $x>a$ ($\rho>1$); inconclusive if $x=a$ ($\rho=1$). At $x=a$, write $a_n=a(1+\varepsilon_n)$ with $\varepsilon_n\to0$, so $b_n=(1+\varepsilon_n)^{-n}$; the outcome then depends on the rate of $\varepsilon_n\to0$ — e.g. $a_n=a+c/n$ gives $b_n\to e^{-c}$ (diverges), while $a_n=a-c/n$ gives $b_n\to e^{c}$ (diverges), so this borderline case generally requires finer analysis of how $a_n$ approaches $a$.


Summary Table

Case$\rho$Conclusion
$a>0,\;x<a$$x/a<1$Converges
$a>0,\;x>a$$x/a>1$Diverges
$a>0,\;x=a$$1$Inconclusive (needs finer analysis)
$a=0$$+\infty$Diverges for all $x>0$

Additional Notes

Result

For $a>0$, the series converges iff $x<a$; for $a=0$, it diverges for all $x>0$:

$$ \rho = \lim_{n\to\infty}\frac{x}{a_n} = \frac{x}{a} \quad (a>0), \qquad \sum_{n=1}^{\infty}\left(\frac{x}{a_n}\right)^n \begin{cases} \text{converges}, & x<a \\ \text{diverges}, & x>a \text{ or } a=0. \end{cases} $$