Using Cauchy's root test, examine the convergence of: sum
Using Cauchy's root test, examine the convergence of: $$ \sum_{n=1}^{\infty} a^{\frac{n^2+1}{n+1}}, \qquad a>0. $$
Explanation of Method
Taking the $n$-th root turns the exponent $\frac{n^2+1}{n+1}$ into $\frac{n^2+1}{n(n+1)}$, and dividing numerator and denominator by $n^2$ shows this exponent tends to $1$. Hence $\sqrt[n]{b_n}\to a^1=a$, so the whole problem reduces to the sign of $a-1$, exactly like a geometric series test.
- Apply the root test to get exponent $\dfrac{n^2+1}{n(n+1)}$.
- Evaluate the limit of this exponent, showing it $\to1$.
- Conclude $\rho=a$ and apply the convergence criterion across all cases of $a$.
Step 1 · Apply the root test
For $a>0$, with $b_n=a^{\frac{n^2+1}{n+1}}$:
$$ \sqrt[n]{b_n} = a^{\frac{n^2+1}{n(n+1)}}. $$
Step 2 · Evaluate the exponent's limit
Dividing numerator and denominator by $n^2$:
$$ \frac{n^2+1}{n^2+n} = \frac{1+\frac1{n^2}}{1+\frac1n} \longrightarrow \frac{1+0}{1+0} = 1. $$
Step 3 · Evaluate $\rho$ and apply the criterion
$$ \rho = \lim_{n\to\infty} a^{\frac{n^2+1}{n(n+1)}} = a^{1} = a. $$
So: if $0<a<1$, $\rho=a<1$ and the series converges; if $a>1$, $\rho=a>1$ and the series diverges; if $a=1$, every term is $1^{\ldots}=1$, so $\sum 1$ trivially diverges; if $a=0$, every term is $0$ and the series trivially converges to $0$.
Summary Table
| Value of $a$ | Conclusion |
|---|---|
| $a=0$ | Trivially converges (sum $=0$) |
| $0<a<1$ | Converges ($\rho=a<1$) |
| $a=1$ | Diverges (terms constant $=1$) |
| $a>1$ | Diverges ($\rho=a>1$) |
Additional Notes
- Unlike most root-test problems where $\rho$ is a fixed number independent of a parameter, here $\rho=a$ directly, so the root test result mirrors exactly the convergence condition of the underlying geometric-type comparison — the exponent's limit of $1$ effectively "reduces" this problem to $\sum a^n$ in disguise.
- The case $a=1$ needed to be checked separately from the root test (which only gives $\rho=1$, inconclusive) by direct substitution — a reminder that whenever $\rho=1$, the terms themselves (not just the root-test limit) must be examined.
- This problem's exponent $\frac{n^2+1}{n+1}\sim n-1$ for large $n$, so $b_n\sim a^{n-1}$, which explains heuristically why the threshold for convergence coincides exactly with the geometric-series threshold $a<1$.
Result
For $a>0$, the series converges if and only if $0<a<1$: