Ratio and Root Tests for Series Convergence

The Essentials

We can apply the ratio and root tests to an infinite series to determine whether it converges or diverges.

Ratio Test:

Given a series \( \Sigma_{n=b}^\infty a_n \), we find the ratio of \( a_{n+1} \) to \( a_n \), take its limit as n goes to infinity, and call this ratio \( \rho \):

\[ \rho = \lim\limits_{n\to\infty} \Big | \frac{a_{n+1}}{a_n} \Big | \]

If \( \rho < 1 \) for a given series, the series converges absolutely.
If \( \rho > 1 \) the series diverges.
if \( \rho = 1 \) the test is inconclusive and we must use another method to test for convergence.

Root Test:

The root test has a similar form to the ratio test. Given a series \( \Sigma_{n=b}^\infty a_n \):

\[ \rho = \lim\limits_{n\to\infty} \big | a_n \big |^{\frac{1}{n}} \]

If \( \rho < 1 \) for a given series, the series converges absolutely.
If \( \rho > 1 \) the series diverges.\\ if \( \rho = 1 \) the test is inconclusive and we must use another method to test for convergence.

Example

These tests are most usefully applied when they simplify the expression \( a_n \) within a series. For example, the series

\[ \Sigma_{n=b}^\infty \frac{(-1)^n(n!)^{2n}}{(2n)!} \]

appears quite complex at first glance. If we were to try to apply another test for convergence, such as the divergence test or a comparison test, we might run into some issues. However, when we apply the root test, the series becomes much simpler:

\[ \rho = \lim\limits_{n\to\infty} \Big | \frac{(-1)^n(n!)^{2n}}{(2n)!} \Big |^{\frac{1}{n}} \longrightarrow \lim\limits_{n\to\infty} \frac{(1)(n!)^2}{(2n!)^{\frac{1}{n}}} \]

Now, with a simplified \( a_n \), we can directly compute the limit:

\[ \lim\limits_{n\to\infty} \frac{(1)(n!)^2}{(2n!)^{\frac{1}{n}}} = \frac{\infty}{\infty^{\frac{1}{\infty}}} = \frac{\infty}{1} = \infty \]

Here, \( \rho > 1 \), so the series diverges.

The ratio test can be similarly applied to simplify a series. We often use the ratio test when we have a series expression that involves factorials and/or differing exponentials, such as:

\[ \Sigma_{n=b}^\infty \frac{9n^2}{(2n+1)!} \]

Applying the ratio test, we get

\[ \rho = \lim\limits_{n\to\infty} \Big | \frac{9(n+1)^2}{(2(n+1)+1)!} \cdot \frac{(2n+1)!}{9n^2} \Big | \]

which we can further simplify:

\[ \rho = \lim\limits_{n\to\infty} \Big | \frac{9(n+1)^2}{9n^2} \cdot \frac{(2n+1)!}{(2n+3)!} \Big | \longrightarrow \lim\limits_{n\to\infty}\frac{9(n+1)^2}{9n^2} \cdot \frac{1}{(2n+3)(2n+2)} \]

From here, we can directly compute the limit. As n goes to infinity, the first part of our expression, \( \frac{9(n+1)^2}{9n^2} \), will go to 1, since \( (n+1) \) is not significantly different from \( n \) when we are working with infinitely large numbers. The second part of our expression, \( \frac{1}{(2n+3)(2n+2)} \) becomes \( \frac{1}{\infty} \) or \( 0 \) when we take its limit. Thus, \( \rho = 1 \cdot 0 = 0 \), which is less than 1, so the series converges absolutely.

Practice

Use either the root or ratio test to determine whether the following series diverge or converge, or state if the test is inconclusive:

  1. \( \Sigma_{n=1}^\infty (1-\frac{1}{n})^{n^2} \)
  2. \( \Sigma_{n=1}^\infty \frac{2^{n^2}}{n^nn!} \)
  3. \( \Sigma_{n=1}^\infty \frac{(-1)^n(n!)^2}{(2n)!} \)

Solutions:

  1. The root test and ratio test are inconclusive.
  2. Diverges by the root test.
  3. Converges by the ratio test.

More Resources