If D'Alembert's test (ratio test) resolves the question
If D'Alembert's test (ratio test) resolves the question of convergence of a series, then Cauchy's test (root test) also resolves it. The converse is not true. Prove this.
Explanation of Method
The forward direction follows from a classical algebra lemma: if $a_{n+1}/a_n\to L$, then $\sqrt[n]{a_n}\to L$ too, obtained by telescoping the ratio into a product and bounding it between $(L-\varepsilon)^{n-N}$ and $(L+\varepsilon)^{n-N}$. The converse fails because $\sqrt[n]{a_n}$ only depends on the overall size of $a_n$, not on the smoothness of consecutive ratios — an explicitly oscillating sequence supplies the counterexample.
- Prove the lemma: $\lim a_{n+1}/a_n=L\implies\lim\sqrt[n]{a_n}=L$.
- Conclude D'Alembert conclusive $\implies$ Cauchy conclusive with the same result.
- Construct a counterexample where Cauchy's test is conclusive but D'Alembert's is not.
Step 1 · Prove $\lim a_{n+1}/a_n=L\implies\lim\sqrt[n]{a_n}=L$
Suppose $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=L$ with $0\le L<\infty$ (the case $L=\infty$ is similar). Given $\varepsilon>0$, choose $N$ so that $L-\varepsilon<\frac{a_{n+1}}{a_n}<L+\varepsilon$ for all $n>N$. Telescoping from $N$ to $n$:
$$ a_n = a_N\prod_{k=N}^{n-1}\frac{a_{k+1}}{a_k} \quad\Longrightarrow\quad (L-\varepsilon)^{n-N} < \frac{a_n}{a_N} < (L+\varepsilon)^{n-N}. $$
Taking $n$-th roots, and using $\sqrt[n]{a_N}\to1$ and $\frac{n-N}n\to1$:
$$ L-\varepsilon \le \liminf_{n\to\infty}\sqrt[n]{a_n} \le \limsup_{n\to\infty}\sqrt[n]{a_n} \le L+\varepsilon. $$
Since $\varepsilon>0$ was arbitrary, $\lim_{n\to\infty}\sqrt[n]{a_n}=L$.
Step 2 · D'Alembert conclusive $\implies$ Cauchy conclusive
If D'Alembert's test gives $L<1$ or $L>1$, Step 1 shows Cauchy's root test gives the identical limit $\rho=L$, so it yields the same convergence/divergence conclusion. Hence whenever D'Alembert's test resolves the series, Cauchy's test resolves it too — with the same answer.
Step 3 · Counterexample for the converse
Let $a_n=\frac1{2^n}$ for even $n$ and $a_n=\frac1{3^n}$ for odd $n$. Then for even $n$, $\frac{a_{n+1}}{a_n}=\frac{1/3^{n+1}}{1/2^n}=\frac13\left(\frac23\right)^n\to0$; for odd $n$, $\frac{a_{n+1}}{a_n}=\frac{1/2^{n+1}}{1/3^n}=\frac12\left(\frac32\right)^n\to\infty$. The ratio oscillates without a limit, so D'Alembert's test is inconclusive. But $\sqrt[n]{a_n}=\frac12$ (even $n$) or $\frac13$ (odd $n$), so $\limsup\sqrt[n]{a_n}=\frac12<1$, and Cauchy's test shows convergence. (Direct check: the series splits into two convergent geometric series.)
Summary Table
| Direction | Status |
|---|---|
| D'Alembert conclusive $\Rightarrow$ Cauchy conclusive | True (proved via telescoping lemma) |
| Cauchy conclusive $\Rightarrow$ D'Alembert conclusive | False (counterexample above) |
Additional Notes
- This confirms the standard hierarchy of convergence tests: Cauchy's root test is strictly stronger than D'Alembert's ratio test, in the sense that it never fails where the ratio test succeeds, and can succeed where the ratio test fails.
- The counterexample's structure — alternating between two distinct geometric rates — is deliberately chosen so the ratio oscillates between $0$ and $\infty$ (no finite limit exists at all), making the failure of D'Alembert's test as dramatic as possible while keeping Cauchy's test trivially easy to apply.
- Problem 79 revisits this exact phenomenon with a more elaborate two-term-per-block sequence, showing the same qualitative behavior (D'Alembert inconclusive, Cauchy conclusive) arises even when the oscillation is milder (ratios $\frac13$ and $2$, rather than $0$ and $\infty$).
Result
D'Alembert's test being conclusive always implies Cauchy's test is conclusive with the same conclusion; the converse fails: