Using Cauchy's root test, examine the convergence of: sum
Using Cauchy's root test, examine the convergence of: $$ \sum_{n=2}^{\infty}\left(\frac{n-2}{n+2}\right)^{n(n+1)}. $$
Explanation of Method
The root test leaves an exponent of $n+1$ rather than $n$, i.e. $\sqrt[n]{b_n}=\left(\frac{n-2}{n+2}\right)^{n+1}$. Rewriting the base as $1-\frac4{n+2}$ turns this into a variant of the classic limit $(1+a/n)^n\to e^a$, giving $\rho=e^{-4}$, comfortably below $1$.
- Apply the root test to reduce the exponent from $n(n+1)$ to $n+1$.
- Rewrite the base as $1-\dfrac4{n+2}$.
- Evaluate the limit via the substitution $m=n+2$ to get $\rho=e^{-4}$.
Step 1 · Apply the root test
With $b_n=\left(\dfrac{n-2}{n+2}\right)^{n(n+1)}$:
$$ \sqrt[n]{b_n} = \left(\frac{n-2}{n+2}\right)^{n+1}. $$
Step 2 · Rewrite the base
$$ \frac{n-2}{n+2} = 1 - \frac4{n+2}. $$
Step 3 · Evaluate the limit
Let $m=n+2$, so $n+1=m-1$:
$$ \left(1-\frac4m\right)^{m-1} = \left[\left(1-\frac4m\right)^{m}\right]^{1-\frac1m}. $$
As $m\to\infty$, $\left(1-\frac4m\right)^m\to e^{-4}$ and $1-\frac1m\to1$, so:
$$ \rho = \lim_{n\to\infty}\left(\frac{n-2}{n+2}\right)^{n+1} = e^{-4} \approx 0.0183. $$
Summary Table
| Quantity | Value |
|---|---|
| $\dfrac{n-2}{n+2}$ (rewritten) | $1-\dfrac4{n+2}$ |
| $\rho=\lim\sqrt[n]{b_n}$ | $e^{-4}\approx0.0183$ |
| Conclusion | Converges |
Additional Notes
- Since $\dfrac{n-2}{n+2}<1$ for every $n\ge2$ and it approaches $1$ from below, the base alone would only barely fail to force convergence via a naive comparison — it's the quadratic exponent $n(n+1)$ that supplies the decisive extra decay, funneling everything into the exponential constant $e^{-4}$.
- The substitution technique here — rewriting an exponent of $n+1$ (or similar) in terms of a shifted variable $m=n+2$ to match the classical form $(1+a/m)^m\to e^a$ — is broadly reusable whenever a fraction close to $1$ is raised to a polynomial power.
- Compare with Problem 76's $\left(a+\frac1n\right)^{-n}$: both problems reduce to evaluating a limit of the form $(1+\text{something}\to0)^{\text{large power}}$, but here the base itself depends on $n$ in a way that produces a genuine $e^{-4}$ rather than a simple reciprocal.
Result
The series converges: