Apply D'Alembert's and Cauchy's tests to the series sum
Apply D'Alembert's and Cauchy's tests to the series $\sum_{n=1}^{\infty} a_n$ where $$ a_{2k-1} = \frac{2^{k-1}}{3^{k-1}}, \qquad a_{2k} = \frac{2^{k-1}}{3^k} \qquad (k=1,2,\ldots), $$ and confirm the second part of the theorem in Problem 78.
Explanation of Method
This series is built from two interleaved geometric-like patterns, so consecutive ratios alternate between a fixed value below $1$ and a fixed value above $1$ — making D'Alembert's test inconclusive by construction. Cauchy's root test, however, only sees the size of each term, and both subsequences of $\sqrt[n]{a_n}$ converge to the same limit $\sqrt{2/3}<1$, resolving the series cleanly.
- Compute the ratio $a_{n+1}/a_n$ separately at odd and even indices.
- Compute $\sqrt[n]{a_n}$ separately at odd and even indices and show both tend to $\sqrt{2/3}$.
- Verify directly by summing the two interleaved geometric series.
Step 1 · D'Alembert's ratio test
At odd-to-even step: $\dfrac{a_{2k}}{a_{2k-1}} = \dfrac{2^{k-1}/3^k}{2^{k-1}/3^{k-1}} = \dfrac13$. At even-to-odd step: $\dfrac{a_{2k+1}}{a_{2k}} = \dfrac{2^k/3^k}{2^{k-1}/3^k} = 2$. So $\dfrac{a_{n+1}}{a_n}$ alternates forever between $\frac13$ and $2$, meaning $\liminf=\frac13\ne2=\limsup$, so the limit does not exist: D'Alembert's test is inconclusive.
Step 2 · Cauchy's root test
For odd $n=2k-1$: $\sqrt[2k-1]{a_{2k-1}} = \left(\frac23\right)^{\frac{k-1}{2k-1}} \to \left(\frac23\right)^{1/2}=\sqrt{2/3}$, since $\frac{k-1}{2k-1}\to\frac12$. For even $n=2k$: $\sqrt[2k]{a_{2k}} = \frac{2^{(k-1)/2k}}{3^{1/2}} \to \frac{2^{1/2}}{3^{1/2}}=\sqrt{2/3}$, since $\frac{k-1}{2k}\to\frac12$. Both subsequences agree, so:
$$ \lim_{n\to\infty}\sqrt[n]{a_n} = \sqrt{\frac23} \approx 0.816 < 1. $$
By Cauchy's root test, the series converges.
Step 3 · Direct verification
Splitting the sum into odd and even parts:
$$ \sum_{n=1}^{\infty} a_n = \sum_{k=1}^\infty\left(\frac23\right)^{k-1} + \frac13\sum_{k=1}^\infty\left(\frac23\right)^{k-1} = \frac43\sum_{k=1}^\infty\left(\frac23\right)^{k-1} = \frac43\cdot3 = 4. $$
Summary Table
| Test | Result |
|---|---|
| D'Alembert (ratio) | Oscillates between $\frac13$ and $2$; inconclusive |
| Cauchy (root) | $\sqrt[n]{a_n}\to\sqrt{2/3}\approx0.816<1$; converges |
| Sum | $4$ |
Additional Notes
- Unlike Problem 78's counterexample, where the ratio oscillated between $0$ and $\infty$, here it oscillates between two finite, nonzero values ($\frac13$ and $2$) straddling $1$ — a milder but equally decisive failure mode for D'Alembert's test.
- The two subsequences of $\sqrt[n]{a_n}$ converging to the exact same limit $\sqrt{2/3}$ is not a coincidence: it reflects that both "halves" of the sequence share the same underlying geometric ratio $2/3$ per full block of two terms, just phase-shifted by one factor of $\frac13$.
- This concrete example gives a fully explicit confirmation of Problem 78's abstract counterexample construction, complete with a closed-form sum ($4$), making it a useful worked illustration of that theorem's second (negative) half.
Result
D'Alembert's test is inconclusive, but Cauchy's test shows convergence: