(ST6) Ratio Test#

In this lesson we are going to see how to:

  • use the Ratio Test to show that a series is convergent or divergent.

Review Video#

The Test#

Ratio Test

Given the series \(\displaystyle \sum_{n=1}^{\infty} a_n\), compute the limit: \(\displaystyle \lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|=L\).

  • If \(L<1\), then the series is convergent (absolutely convergent).

  • If \(L>1\) or is infinite, then the series is divergent.

  • If \(L=1\), then this test is inconclusive.

Example 1#

Determine if the following series is convergent or divergent.

\[ \sum_{n=1}^{\infty} (-1)^{n-1}\dfrac{n^5}{2^n} \]

(Click to see the steps.)

We start by setting up the limit we need to calculate.

When we do this it is helpful to think of the division by \(a_n\) more as multiplication by the reciprocal of \(a_n\).

\[\begin{split} \lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right| & = \lim_{n\to\infty}\left|a_{n+1}\cdot \dfrac{1}{a_n}\right| \\[10pt] & = \lim_{n\to\infty}\left|\dfrac{(-1)^{n}(n+1)^5}{2^{n+1}}\cdot \dfrac{2^n}{(-1)^{n-1}n^5}\right| \\[10pt] \end{split}\]

The next step is to group like terms in their own separate fractions:

\[\begin{split} = \lim_{n\to\infty}\left|\dfrac{(-1)^{n}}{(-1)^{n-1}}\cdot \dfrac{(n+1)^5}{n^5}\cdot \dfrac{2^n}{2^{n+1}}\right| \\[10pt] \end{split}\]

Now we can simplify each fraction.

Since we are working with terms inside of an absolute value sign, the powers of \((-1)\) will cancel out completely. And everything else is positive, so once we cancel the \((-1)\) terms, we can drop the absolute value sign.

\[ = \lim_{n\to\infty}\dfrac{(n+1)^5}{n^5}\cdot \dfrac{2^n}{2^{n+1}} \]

Simplify the powers of \(2\) in the last fraction:

\[ = \lim_{n\to\infty}\dfrac{(n+1)^5}{n^5}\cdot \dfrac{1}{2} \]

And use an exponent law to “factor off” the exponent of \(5\) in the first fraction:

\[ = \lim_{n\to\infty}\left(\dfrac{n+1}{n}\right)^5 \cdot \dfrac{1}{2} \]

Now we are ready to actually calculate the limit:

\[ \lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right| =\lim_{n\to\infty}\left(\dfrac{n+1}{n}\right)^5 \cdot \dfrac{1}{2} \]

And since the power function \((\cdot)^5\) is continuous, we can bring the limit inside the parentheses:

\[ =\left(\lim_{n\to\infty}\dfrac{n+1}{n}\right)^5 \cdot \dfrac{1}{2} \]

Do the division:

\[ =\left(\lim_{n\to\infty} 1+\dfrac{1}{n}\right)^5 \cdot \dfrac{1}{2} \]

And calculate:

\[\begin{split} &=\left( 1+0\right)^5 \cdot \dfrac{1}{2}\\[10pt] &=\dfrac{1}{2}\\[10pt] \end{split}\]

Since the limit here is \(L=\tfrac{1}{2}\) which is less than \(1\), we can conclude by the ratio test that the series is convergent.

Example 2#

Determine if the following series is convergent or divergent.

\[ \sum_{n=1}^{\infty} \dfrac{n!}{3^n \sqrt{n}} \]

(Click to see the steps.)

We start by setting up the limit we need to calculate.

When we do this it is helpful to think of the division by \(a_n\) more as multiplication by the reciprocal of \(a_n\).

\[\begin{split} \lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right| & = \lim_{n\to\infty}\left|a_{n+1}\cdot \dfrac{1}{a_n}\right| \\[10pt] & = \lim_{n\to\infty}\left|\dfrac{(n+1)!}{3^{n+1} \sqrt{n+1}}\cdot \dfrac{3^n \sqrt{n}}{n!}\right| \\[10pt] \end{split}\]

The next step is to group like terms in their own separate fractions:

\[\begin{split} = \lim_{n\to\infty}\left|\dfrac{(n+1)!}{n!}\cdot \dfrac{3^n}{3^{n+1}}\cdot \dfrac{\sqrt{n}}{\sqrt{n+1}}\right| \\[10pt] \end{split}\]

Now we can simplify each fraction.

We’ll start by simplifying the powers of \(3\) in the middle fraction, and while we’re at it: since everything is positive we can drop the absolute value sign.

\[\begin{split} = \lim_{n\to\infty}\dfrac{(n+1)!}{n!}\cdot \dfrac{1}{3}\cdot \dfrac{\sqrt{n}}{\sqrt{n+1}} \\[10pt] \end{split}\]

Simplify the factorials. Remember that \((n+1)! = (n+1)\, n!\)

\[\begin{split} &= \lim_{n\to\infty}\dfrac{(n+1)\, n!}{n!}\cdot \dfrac{1}{3}\cdot \dfrac{\sqrt{n}}{\sqrt{n+1}} \\[10pt] &= \lim_{n\to\infty}(n+1)\cdot \dfrac{1}{3}\cdot \dfrac{\sqrt{n}}{\sqrt{n+1}} \\[10pt] \end{split}\]

And use an exponent law to “factor off” the square root in the last fraction:

\[\begin{split} = \lim_{n\to\infty}(n+1)\cdot \dfrac{1}{3}\cdot \sqrt{\dfrac{n}{n+1}} \\[10pt] \end{split}\]

Now we are ready to actually calculate the limit:

\[\begin{split} \lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}(n+1)\cdot \dfrac{1}{3}\cdot \sqrt{\dfrac{n}{n+1}} \\[10pt] \end{split}\]

We first note that the limit of the square root term is equal to \(1\), since:

\[ \lim_{n\to\infty} \sqrt{\dfrac{n}{n+1}} = \sqrt{\lim_{n\to\infty} \dfrac{n}{n+1}} = \sqrt{\lim_{n\to\infty} \dfrac{1}{1+\tfrac{1}{n}}}=\sqrt{\dfrac{1}{1+0}}=1 \]

This means our original limit is infinite since the \((n+1)\) term goes to infinity while the other terms approach a non-zero constant. Therefore:

\[ \lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}(n+1)\cdot \dfrac{1}{3}\cdot \sqrt{\dfrac{n}{n+1}} =\infty \]

Since the limit \(L\) here is infinite, we can conclude by the ratio test that the series is divergent.

Inconclusive#

Remember that the Ratio Test is inconclusive when \(L=1\). This means the series might converge or it might diverge, we just don’t know; the test is inconclusive.

In the case of a \(p\)-series, the Ratio Test is always inconclusive. For example:

\(\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n}\) is a divergent \(p\)-series with \(p=1\leq1\) but the ratio test is inconclusive.

The Ratio Test is inconclusive since:

\[\begin{split} \lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right| & = \lim_{n\to\infty}\left|a_{n+1}\cdot \dfrac{1}{a_n}\right| \\[10pt] & = \lim_{n\to\infty}\left|\dfrac{1}{n+1}\cdot \dfrac{n}{1}\right| \\[10pt] & = \lim_{n\to\infty}\dfrac{n}{n+1} \\[10pt] & = \lim_{n\to\infty}\dfrac{\tfrac{n}{n}}{\tfrac{n}{n}+\tfrac{1}{n}}\\[10pt] & = \lim_{n\to\infty}\dfrac{1}{1+\tfrac{1}{n}}\\[10pt] & = \dfrac{1}{1+0} \\[10pt] & = 1 \\[10pt] \end{split}\]
\(\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^2}\) is a convergent \(p\)-series with \(p=2>1\) but the ratio test is inconclusive.

The Ratio Test is inconclusive since:

\[\begin{split} \lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right| & = \lim_{n\to\infty}\left|a_{n+1}\cdot \dfrac{1}{a_n}\right| \\[10pt] & = \lim_{n\to\infty}\left|\dfrac{1}{(n+1)^2}\cdot \dfrac{n^2}{1}\right| \\[10pt] & = \lim_{n\to\infty}\dfrac{n^2}{(n+1)^2} \\[10pt] & = \lim_{n\to\infty}\left(\dfrac{n}{n+1}\right)^2 \\[10pt] & = \left(\lim_{n\to\infty}\dfrac{n}{n+1}\right)^2 \\[10pt] & = \left(\lim_{n\to\infty}\dfrac{1}{1+\tfrac{1}{n}}\right)^2 \\[10pt] & = \left(\dfrac{1}{1+0}\right)^2 \\[10pt] & = 1 \\[10pt] \end{split}\]