Root Test

Approach: The general term is built from a power of $n$, so the root test applies directly: take the $n$-th root of $a_n$ and compare the limit with $1$.

Steps:

1. Let $a_n = \frac{1}{n^n}$, then $\sqrt[n]{a_n} = \frac{1}{n}$
2. Take the limit: $\rho = \lim_{n \to \infty} \frac{1}{n} = 0$
3. Compare with $1$: $\rho = 0 < 1$
4. By the root test, the series converges

Answer: The series converges ($\rho = 0 < 1$).

Approach: Numerator and denominator both contain powers of $n$; after taking the $n$-th root each factor tends to a constant, so compute those two limits first and then compare the result with $1$.

Steps:

1. Let $a_n = \frac{n^2}{3^n}$, then $\sqrt[n]{a_n} = \frac{n^{2/n}}{3}$
2. Limit of the numerator: $n^{2/n} = e^{\frac{2\ln n}{n}} \to e^0 = 1$, because $\frac{\ln n}{n} \to 0$
3. Hence $\rho = \lim_{n \to \infty} \frac{n^{2/n}}{3} = \frac{1}{3}$
4. Compare with $1$: $\rho = \frac{1}{3} < 1$, so the series converges by the root test

Answer: The series converges ($\rho = \frac{1}{3} < 1$).

Approach: The $n$-th root of the exponential term $2^n$ is still $2$, while the $n$-th root of the polynomial term $n^2$ tends to $1$; the quotient of the two is therefore larger than $1$.

Steps:

1. Let $a_n = \frac{2^n}{n^2}$, then $\sqrt[n]{a_n} = \frac{2}{n^{2/n}}$
2. Limit of the denominator: $n^{2/n} = e^{\frac{2\ln n}{n}} \to 1$
3. Hence $\rho = \lim_{n \to \infty} \frac{2}{n^{2/n}} = 2$
4. Compare with $1$: $\rho = 2 > 1$, so the series diverges by the root test (indeed $a_n = \frac{2^n}{n^2}$ does not even tend to $0$)

Answer: The series diverges ($\rho = 2 > 1$).

Approach: First run the root test and compute $\rho$. If the result is $\rho = 1$, the test is inconclusive and a different test has to be used.

Steps:

1. Let $a_n = \frac{1}{n^p}$, then $\sqrt[n]{a_n} = n^{-\frac{p}{n}} = e^{-\frac{p\ln n}{n}}$
2. Because $\frac{\ln n}{n} \to 0$, we get $\rho = \lim_{n \to \infty} e^{-\frac{p\ln n}{n}} = e^0 = 1$
3. The case $\rho = 1$ is exactly the case in which the root test fails: it gives no verdict on convergence
4. Another test must be chosen: the $p$-series test gives convergence for $p > 1$ and divergence for $0 < p \leq 1$

Answer: $\rho = 1$, so the root test is inconclusive and a different test is needed; with the $p$-series test the series converges for $p > 1$ and diverges for $0 < p \leq 1$.