The Root Test for Positive Series Examples 1

If the sequence of terms $\{ a_n \}$ is ultimately positive and that $\lim_{n \to \infty} (a_n)^{1/n} = L$ where $L$ is some nonnegative real number or $\infty$, then recall from The Root Test for Positive Series page that if $0 ≤ L < 1$ then the series $\sum_{n=1}^{\infty} a_n$ converges. If $1 < L ≤ \infty$ then the series $\sum_{n=1}^{\infty} a_n$ diverges to infinity. If $L = 1$ then this test does not provide any additional information about the convergence/divergence of the series $\sum_{n=1}^{\infty} a_n$.

We are now going to look at some examples applying the root test.

Example 1

Using the root test, determine whether the series $\sum_{n=1}^{\infty} \left ( \frac{n^2 + 3n}{n^4 + 1} \right )^n$ is convergent or divergent.

Applying the root test directly we get that:

(1)

\begin{align} \lim_{n \to \infty} \left ( \left ( \frac{n^2 + 3n}{n^4 + 1} \right )^n \right)^{1/n} = \lim_{n \to \infty} \frac{n^2 + 3n}{n^4 + 1} = \frac{\frac{1}{n^2} + \frac{3}{n^3}}{1 + \frac{1}{n^4}} = 0 \end{align}

Since $L = 0$ and $0 ≤ L < 1$ we conclude that this series is convergent.

Example 2

Using the root test, determine whether the series $\sum_{n=1}^{\infty} \left ( 1 - \frac{1}{n} \right )^{n^2}$ is convergent or divergent.

Applying the root test directly we get that:

(2)

\begin{align} \lim_{n \to \infty} \left ( \left ( 1 - \frac{1}{n} \right )^{n^2} \right )^{1/n} = \lim_{n \to \infty} \left( 1 - \frac{1}{n} \right)^n \end{align}

Now we will solve this limit using the natural logarithm. Let $m = \left( 1 - \frac{1}{n} \right)^n$. Then $\ln (m) = \ln \left ( \left( 1 - \frac{1}{n} \right)^n \right) = n \ln \left ( 1 - \frac{1}{n} \right)$

And so:

(3)

\begin{align} \lim_{n \to \infty} \left( 1 - \frac{1}{n} \right)^n = \lim_{n \to \infty} e^{n \ln (1 - \frac{1}{n})} = e^{\lim_{n \to \infty} \frac{\ln (1 - 1/n)}{1/n}} \end{align}

Now let $f(x) = \frac{\ln (1 - 1/x)}{1/x}$. By L'Hospital's rule, we have that $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} f'(x)$ since $\lim_{n \to \infty} \ln ( 1 - 1/x) = \ln 1 = 0$ and $\lim_{n \to \infty} \frac{1}{n} = 0$, so we have an indeterminate form $0 / 0$ and so:

(4)

\begin{align} \lim_{x \to \infty} \frac{\frac{1}{1 - 1/x} \cdot \frac{1}{x^2}}{-\frac{1}{x^2}} = \lim_{x \to \infty} -\left (1 - \frac{1}{x} \right) = -1 \end{align}

And so $\lim_{n \to \infty} \lim_{n \to \infty} \frac{\ln (1 - 1/n)}{1/n} = -1$. Therefore:

(5)

\begin{align} \lim_{n \to \infty} \left ( \left ( 1 - \frac{1}{n} \right )^{n^2} \right )^{1/n} = e^{\lim_{n \to \infty} \frac{\ln (1 - 1/n)}{1/n}} = e^{-1} = \frac{1}{e} \end{align}

Since $L = \frac{1}{e}$ and $0 ≤ L < 1$, then by the root test this series is convergent.

Example 3

Using the root test, determine whether the series $\sum_{n=1}^{\infty} \frac{n^n}{2^{1+2n}}$ is convergent or divergent.

Applying the root test we get that:

(6)

\begin{align} \lim_{n \to \infty} \left ( \frac{n^n}{2^{1+2n}} \right )^{1/n} = \lim_{n \to \infty} \frac{n}{2^{\frac{1 + 2n}{n}}} = \lim_{n \to \infty} \frac{n}{2^{1/n + 2}} = \infty \end{align}

Since $L = \infty$ and $1 < L ≤ \infty$ then by the root test this series is divergent.