If varlimsup_(n to infinity) nth root of (a_n) = q where
If $\varlimsup_{n\to\infty} \sqrt[n]{a_n} = q$ where $a_n\ge0$, then for $q<1$ the series is convergent and for $q>1$ the series is divergent. Prove this.
Explanation of Method
For $q<1$, the definition of $\limsup$ guarantees $\sqrt[n]{a_n}$ is eventually trapped below some $r<1$, so $a_n<r^n$ eventually and comparison with a convergent geometric series finishes the proof. For $q>1$, the definition guarantees infinitely many $n$ with $\sqrt[n]{a_n}$ above some $s>1$, forcing $a_n\not\to0$ infinitely often — violating the necessary condition for convergence.
- Recall the $\varepsilon$-definition of $\limsup$.
- Prove convergence for $q<1$ via comparison with a geometric series.
- Prove divergence for $q>1$ via the necessary condition $a_n\to0$.
Step 1 · Recall the definition of $\limsup$
$q=\varlimsup_{n\to\infty}\sqrt[n]{a_n}$ means: for every $\varepsilon>0$, $\sqrt[n]{a_n}<q+\varepsilon$ for all sufficiently large $n$; and for every $\varepsilon>0$, $\sqrt[n]{a_n}>q-\varepsilon$ for infinitely many $n$.
Step 2 · Convergence for $q<1$
Choose $r$ with $q<r<1$, and let $\varepsilon=r-q>0$. There exists $N$ such that $\sqrt[n]{a_n}<q+\varepsilon=r$ for all $n>N$, i.e. $a_n<r^n$. Since $0<r<1$, $\sum r^n$ converges (geometric series), so by comparison $\sum_{n>N}a_n$ converges, and adding the finitely many terms $a_1,\ldots,a_N$ preserves convergence: $\sum_{n=1}^\infty a_n$ converges.
Step 3 · Divergence for $q>1$
Choose $s$ with $1<s<q$, and let $\varepsilon=q-s>0$. There are infinitely many $n$ with $\sqrt[n]{a_n}>q-\varepsilon=s$, i.e. $a_n>s^n$ for infinitely many $n$. Since $s>1$, $s^n\to\infty$, so along this subsequence $a_{n_k}\to\infty$; in particular $a_n\not\to0$. Since $a_n\to0$ is necessary for $\sum a_n$ to converge, the series diverges.
Summary Table
| Value of $q$ | Conclusion |
|---|---|
| $q<1$ | Converges (comparison with $\sum r^n$, $q<r<1$) |
| $q>1$ | Diverges ($a_n\not\to0$ along a subsequence) |
| $q=1$ | Inconclusive |
Additional Notes
- The case $q=1$ is genuinely ambiguous: $\sum\frac1n$ has $\sqrt[n]{1/n}=1/n^{1/n}\to1$ and diverges, while $\sum\frac1{n^2}$ has $\sqrt[n]{1/n^2}=1/n^{2/n}\to1$ and converges — identical root-test limits, opposite outcomes.
- This proof is the formal justification underlying every root-test computation in Problems 58–77: whenever those problems computed $\rho=\limsup\sqrt[n]{b_n}$ and compared it to $1$, they were implicitly invoking exactly this theorem.
- The two halves of the proof use genuinely different tools — comparison with a convergent geometric series for $q<1$, versus the necessary condition $a_n\to0$ for $q>1$ — reflecting the asymmetry between proving convergence (a "constructive" bound) and proving divergence (a single failure suffices).
Result
Cauchy's root test is proved in full: