Examine the convergence of the series: sum from n=1
Examine the convergence of the series: $$ \sum_{n=1}^{\infty} 5^{-\sqrt[3]{n}}. $$
Explanation of Method
The terms $a_n=5^{-n^{1/3}}$ decrease monotonically, but neither the ratio nor the root test applied directly to $a_n$ gives a clean limit, since $n^{1/3}$ is neither linear nor bounded. Cauchy's condensation test (Problem 98) converts this into the condensed series $\sum2^ka_{2^k}=\sum2^k/5^{2^{k/3}}$, whose root test limit is trivially $0$, since the doubly-exponential denominator $5^{2^{k/3}}$ overwhelms $2^k$.
- Verify $a_n=5^{-n^{1/3}}$ is positive and decreasing.
- Form the condensed series $2^ka_{2^k}=2^k/5^{2^{k/3}}$.
- Apply the root test to the condensed series, showing it converges.
Step 1 · Verify the hypotheses of condensation
$f(x)=5^{-x^{1/3}}$ has $f'(x)=5^{-x^{1/3}}\cdot(-\ln5)\cdot\frac13x^{-2/3}<0$ for $x>0$, so $a_n$ is positive and decreasing — condensation applies.
Step 2 · Form the condensed series
$$ 2^k a_{2^k} = 2^k\cdot5^{-(2^k)^{1/3}} = \frac{2^k}{5^{2^{k/3}}}. $$
Step 3 · Apply the root test to the condensed series
With $b_k=2^k/5^{2^{k/3}}$: $\sqrt[k]{b_k}=\dfrac2{5^{2^{k/3}/k}}$. Since $2^{k/3}$ grows exponentially while $k$ grows linearly, $2^{k/3}/k\to\infty$, so $5^{2^{k/3}/k}\to\infty$ and $\sqrt[k]{b_k}\to0<1$. By the root test, $\sum b_k$ converges, so by Cauchy's condensation test, the original series converges too.
Summary Table
| Quantity | Value |
|---|---|
| Condensed term | $2^k/5^{2^{k/3}}$ |
| $\lim\sqrt[k]{b_k}$ | $0$ |
| Conclusion | Converges |
Additional Notes
- This is a textbook use case for Cauchy's condensation test: the original terms decay at a rate ($n^{-1/3}$ in the exponent) too irregular for a direct ratio or root test to handle cleanly, but the condensed sequence's doubly-exponential structure makes the root test trivial.
- The key structural fact is that $5^{-n^{1/3}}$ decays slower than any fixed exponential $r^{-n}$ but still fast enough to be summable — a decay rate "between" polynomial and exponential, exactly the kind of borderline case condensation is designed to resolve.
- The same technique would apply to any $5^{-n^\beta}$ with $\beta>0$: condensation always produces $2^k/5^{2^{k\beta}}$, whose root test limit is $0$ regardless of $\beta$, so $\sum5^{-n^\beta}$ converges for every $\beta>0$.
Result
The series converges: