Using Cauchy's root test, examine the convergence of: sum
Using Cauchy's root test, examine the convergence of: $$ \sum_{n=1}^{\infty} \frac{n^k}{a^n+b^n}, \qquad a>1,\; b>1. $$
Explanation of Method
The key step is evaluating $\sqrt[n]{a^n+b^n}$: factoring out the larger of $a,b$ shows this tends to $\max\{a,b\}$, since the smaller term's contribution vanishes after taking the $n$-th root. Combined with $\sqrt[n]{n^k}\to1$ for any fixed $k$, this yields $\rho=1/\max\{a,b\}<1$ unconditionally.
- Apply the root test, separating $\sqrt[n]{n^k}$ from $\sqrt[n]{a^n+b^n}$.
- Evaluate $\sqrt[n]{n^k}\to1$.
- Evaluate $\sqrt[n]{a^n+b^n}\to\max\{a,b\}$ by factoring, then conclude.
Step 1 · Apply the root test
With $c_n=\dfrac{n^k}{a^n+b^n}$:
$$ \sqrt[n]{c_n} = \frac{n^{k/n}}{\sqrt[n]{a^n+b^n}}. $$
Step 2 · Evaluate $\lim n^{k/n}$
For any fixed $k$: $\;n^{k/n}=(n^{1/n})^k\to1^k=1$.
Step 3 · Evaluate $\lim\sqrt[n]{a^n+b^n}$ and conclude
Assume WLOG $a\ge b>1$. Factor out $a^n$:
$$ \sqrt[n]{a^n+b^n} = a\cdot\sqrt[n]{1+\left(\frac ba\right)^n}. $$
If $a>b$, $\left(\frac ba\right)^n\to0$, so $\sqrt[n]{1+(b/a)^n}\to1$; if $a=b$, $\sqrt[n]{2a^n}=\sqrt[n]2\cdot a\to a$. Either way, $\sqrt[n]{a^n+b^n}\to\max\{a,b\}$. Therefore:
$$ \rho = \frac{1}{\max\{a,b\}} < 1, $$
since $a,b>1$. By Cauchy's root test, the series converges for every $k$.
Summary Table
| Quantity | Value |
|---|---|
| $\lim n^{k/n}$ | $1$, any fixed $k$ |
| $\lim\sqrt[n]{a^n+b^n}$ | $\max\{a,b\}$ |
| $\rho$ | $1/\max\{a,b\}$ |
| Conclusion | Converges for all $k$, $a,b>1$ |
Additional Notes
- The fact $\sqrt[n]{a^n+b^n}\to\max\{a,b\}$ is a useful general lemma: it extends naturally to $\sqrt[n]{a_1^n+\cdots+a_m^n}\to\max\{a_1,\ldots,a_m\}$ for any finite collection of positive constants, by the same factoring argument.
- This problem's convergence is independent of $k$ — no matter how large the polynomial power in the numerator, the exponential denominator $a^n+b^n$ always wins, matching the general principle that exponential growth eventually dominates any fixed polynomial growth (already seen in Problem 70's $n/a^n$).
- Setting $b\to0^+$ (outside the stated hypothesis $b>1$, but instructive) would recover exactly Problem 76's structure with $a$ playing the same role, reinforcing that $\max\{a,b\}$ is the natural generalization of a single base $a$.
Result
The series converges for all $a>1$, $b>1$, and every value of $k$: