Using Cauchy's root test, examine the convergence of: sum
Using Cauchy's root test, examine the convergence of: $$ \sum_{n=1}^{\infty} \frac{n^3}{\left(a+\frac{1}{n}\right)^n}, \qquad a>0. $$
Explanation of Method
The root test gives $\sqrt[n]{b_n}=\dfrac{n^{3/n}}{a+1/n}$. The numerator's limit $n^{3/n}\to1$ is computed via logarithms and L'Hôpital's rule, while the denominator simply tends to $a$, so $\rho=1/a$ — reducing the whole problem to comparing $a$ against $1$, with the boundary case $a=1$ requiring a direct check that the terms don't vanish.
- Apply the root test and split into $n^{3/n}$ and $a+1/n$.
- Evaluate $\lim n^{3/n}=1$ via logarithms.
- Combine to get $\rho=1/a$, and resolve $a=1$ by direct substitution.
Step 1 · Apply the root test
With $b_n=\dfrac{n^3}{\left(a+\frac1n\right)^n}$:
$$ \sqrt[n]{b_n} = \frac{\sqrt[n]{n^3}}{a+\frac1n} = \frac{n^{3/n}}{a+\frac1n}. $$
Step 2 · Evaluate $\lim_{n\to\infty} n^{3/n}$
Let $L=\lim n^{3/n}$; taking logs and applying L'Hôpital's rule:
$$ \ln L = \lim_{n\to\infty}\frac{3\ln n}{n} \overset{\text{L'H}}{=} \lim_{n\to\infty}\frac{3/n}{1} = 0 \quad\Longrightarrow\quad L=1. $$
Step 3 · Combine and resolve $a=1$
Since $a+\frac1n\to a$:
$$ \rho = \frac{\lim n^{3/n}}{\lim(a+\frac1n)} = \frac1a. $$
So: $a>1\Rightarrow\rho<1$, converges; $a<1\Rightarrow\rho>1$, diverges; $a=1\Rightarrow\rho=1$, inconclusive by the test alone. Directly, at $a=1$: $\left(1+\frac1n\right)^n\to e$, so $b_n=\frac{n^3}{(1+1/n)^n}\to\frac{n^3}{e}\to\infty\ne0$, so the series diverges at $a=1$ too.
Summary Table
| Value of $a$ | $\rho$ | Conclusion |
|---|---|---|
| $a>1$ | $1/a<1$ | Converges |
| $a=1$ | $1$ (inconclusive by root test) | Diverges (terms $\to\infty$) |
| $0<a<1$ | $1/a>1$ | Diverges |
Additional Notes
- This is a close cousin of Problem 70's $\sum n/a^n$, but the polynomial factor is $n^3$ instead of $n$ and the base is perturbed by $+1/n$ — neither change affects $\rho$, since $n^{k/n}\to1$ for any fixed $k$ and the perturbation $1/n\to0$ vanishes in the limit.
- The $a=1$ boundary is a good illustration of why root-test inconclusiveness ($\rho=1$) must always be followed up by checking whether the necessary condition $b_n\to0$ even holds — here it dramatically fails, since $b_n\to\infty$.
- More generally, for any fixed polynomial factor $n^k$ in the numerator and any base tending to $a>0$ in the denominator, the same pattern $\rho=1/a$ emerges — a reusable template for the whole family of problems like 76, 77.
Result
For $a>0$, the series converges iff $a>1$: