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### Section 4-10 : Ratio Test

In this section we are going to take a look at a test that we can use to see if a series is absolutely convergent or not. Recall that if a series is absolutely convergent then we will also know that it’s convergent and so we will often use it to simply determine the convergence of a series.

Before proceeding with the test let’s do a quick reminder of factorials. This test will be particularly useful for series that contain factorials (and we will see some in the applications) so let’s make sure we can deal with them before we run into them in an example.

If \(n\) is an integer such that \(n \ge 0\) then \(n\) factorial is defined as,

\[\begin{align*}n! & = n\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( 3 \right)\left( 2 \right)\left( 1 \right) & \hspace{0.15in} & {\mbox{if }}n \ge 1\\ 0! & = 1 & \hspace{0.15in} & {\mbox{by definition}}\end{align*}\]Let’s compute a couple real quick.

\[\begin{align*}& 1! = 1\\ & 2! = 2\left( 1 \right) = 2\\ & 3! = 3\left( 2 \right)\left( 1 \right) = 6\\ & 4! = 4\left( 3 \right)\left( 2 \right)\left( 1 \right) = 24\\ & 5! = 5\left( 4 \right)\left( 3 \right)\left( 2 \right)\left( 1 \right) = 120\end{align*}\]In the last computation above, notice that we could rewrite the factorial in a couple of different ways. For instance,

\[\begin{align*}5! & = 5\underbrace {\left( 4 \right)\left( 3 \right)\left( 2 \right)\left( 1 \right)}_{4!} = 5 \cdot 4!\\ 5! & = 5\left( 4 \right)\underbrace {\left( 3 \right)\left( 2 \right)\left( 1 \right)}_{3!} = 5\left( 4 \right) \cdot 3!\end{align*}\]In general, we can always “strip out” terms from a factorial as follows.

\[\begin{align*}n! & = n\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( {n - k} \right)\left( {n - \left( {k + 1} \right)} \right) \cdots \left( 3 \right)\left( 2 \right)\left( 1 \right)\\ & = n\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( {n - k} \right) \cdot \left( {n - \left( {k + 1} \right)} \right)!\\ & = n\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( {n - k} \right) \cdot \left( {n - k - 1} \right)!\end{align*}\]We will need to do this on occasion so don’t forget about it.

Also, when dealing with factorials we need to be very careful with parenthesis. For instance, \(\left( {2n} \right)! \ne 2\,\,n!\) as we can see if we write each of the following factorials out.

\[\begin{align*}\left( {2n} \right)! & = \left( {2n} \right)\left( {2n - 1} \right)\left( {2n - 2} \right) \cdots \left( 3 \right)\left( 2 \right)\left( 1 \right)\\ 2\,\,n! & = 2\left[ {\left( n \right)\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( 3 \right)\left( 2 \right)\left( 1 \right)} \right]\end{align*}\]Again, we will run across factorials with parenthesis so don’t drop them. This is often one of the more common mistakes that students make when they first run across factorials.

Okay, we are now ready for the test.

#### Ratio Test

Suppose we have the series \(\displaystyle \sum {{a_n}} \). Define,

\[L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right|\]Then,

- if \(L < 1\) the series is absolutely convergent (and hence convergent).
- if \(L > 1\) the series is divergent.
- if \(L = 1\) the series may be divergent, conditionally convergent, or absolutely convergent.

A proof of this test is at the end of the section.

Notice that in the case of \(L = 1\) the ratio test is pretty much worthless and we would need to resort to a different test to determine the convergence of the series.

Also, the absolute value bars in the definition of \(L\) are absolutely required. If they are not there it will be impossible for us to get the correct answer.

Let’s take a look at some examples.

With this first example let’s be a little careful and make sure that we have everything down correctly. Here are the series terms \({a_n}\).

\[{a_n} = \frac{{{{\left( { - 10} \right)}^n}}}{{{4^{2n + 1}}\left( {n + 1} \right)}}\]Recall that to compute \({a_{n + 1}}\) all that we need to do is substitute *n+1* for all the \(n\)’s in \({a_n}\).

Now, to define \(L\) we will use,

\[L = \mathop {\lim }\limits_{n \to \infty } \left| {{a_{n + 1}} \cdot \frac{1}{{{a_n}}}} \right|\]since this will be a little easier when dealing with fractions as we’ve got here. So,

\[\begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( { - 10} \right)}^{n + 1}}}}{{{4^{2n + 3}}\left( {n + 2} \right)}}\,\,\frac{{{4^{2n + 1}}\left( {n + 1} \right)}}{{{{\left( { - 10} \right)}^n}}}} \right|\\ & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{ - 10\left( {n + 1} \right)}}{{{4^2}\left( {n + 2} \right)}}} \right|\\ & = \frac{{10}}{{16}}\mathop {\lim }\limits_{n \to \infty } \frac{{n + 1}}{{n + 2}}\\ & = \frac{{10}}{{16}} < 1\end{align*}\]So, \(L < 1\) and so by the Ratio Test the series converges absolutely and hence will converge.

As seen in the previous example there is usually a lot of canceling that will happen in these. Make sure that you do this canceling. If you don’t do this kind of canceling it can make the limit fairly difficult.

Now that we’ve worked one in detail we won’t go into quite the detail with the rest of these. Here is the limit.

\[L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\left( {n + 1} \right)!}}{{{5^{n + 1}}}}\,\frac{{{5^n}}}{{n!}}} \right| = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {n + 1} \right)!}}{{5\,\,n!}}\]In order to do this limit we will need to eliminate the factorials. We simply can’t do the limit with the factorials in it. To eliminate the factorials we will recall from our discussion on factorials above that we can always “strip out” terms from a factorial. If we do that with the numerator (in this case because it’s the larger of the two) we get,

\[L = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {n + 1} \right)\,\,n!}}{{5\,\,n!}}\]at which point we can cancel the \(n\)! for the numerator an denominator to get,

\[L = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {n + 1} \right)}}{5} = \infty > 1\]So, by the Ratio Test this series diverges.

In this case be careful in dealing with the factorials.

\[\begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( {n + 1} \right)}^2}}}{{\left( {2\left( {n + 1} \right) - 1} \right)!}}\,\frac{{\left( {2n - 1} \right)!}}{{{n^2}}}} \right|\\ & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( {n + 1} \right)}^2}}}{{\left( {2n + 1} \right)!}}\,\frac{{\left( {2n - 1} \right)!}}{{{n^2}}}} \right|\\ & = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {n + 1} \right)}^2}}}{{\left( {2n + 1} \right)\left( {2n} \right)\left( {2n - 1} \right)!}}\,\frac{{\left( {2n - 1} \right)!}}{{{n^2}}}\\ & = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {n + 1} \right)}^2}}}{{\left( {2n + 1} \right)\left( {2n} \right)\left( {{n^2}} \right)}}\\ & = 0 < 1\end{align*}\]So, by the Ratio Test this series converges absolutely and so converges.

Do not mistake this for a geometric series. The \(n\) in the denominator means that this isn’t a geometric series. So, let’s compute the limit.

\[\begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{9^{n + 1}}}}{{{{\left( { - 2} \right)}^{n + 2}}\,\left( {n + 1} \right)}}\,\,\frac{{{{\left( { - 2} \right)}^{n + 1}}\,\,n}}{{{9^n}}}} \right|\\ & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{9\,n}}{{\left( { - 2} \right)\,\left( {n + 1} \right)}}} \right|\\ & = \frac{9}{2}\mathop {\lim }\limits_{n \to \infty } \frac{n}{{n + 1}}\\ & = \frac{9}{2} > 1\end{align*}\]Therefore, by the Ratio Test this series is divergent.

In the previous example the absolute value bars were required to get the correct answer. If we hadn’t used them we would have gotten \(L = - \frac{9}{2} < 1\) which would have implied a convergent series!

Now, let’s take a look at a couple of examples to see what happens when we get\(L = 1\). Recall that the ratio test will not tell us anything about the convergence of these series. In both of these examples we will first verify that we get \(L = 1\) and then use other tests to determine the convergence.

Let’s first get \(L\).

\[L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{{{\left( {n + 1} \right)}^2} + 1}}\,\,\frac{{{n^2} + 1}}{{{{\left( { - 1} \right)}^n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^2} + 1}}{{{{\left( {n + 1} \right)}^2} + 1}} = 1\]So, as implied earlier we get \(L = 1\) which means the ratio test is no good for determining the convergence of this series. We will need to resort to another test for this series. This series is an alternating series and so let’s check the two conditions from that test.

\[\mathop {\lim }\limits_{n \to \infty } {b_n} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^2} + 1}} = 0\] \[{b_n} = \frac{1}{{{n^2} + 1}} > \frac{1}{{{{\left( {n + 1} \right)}^2} + 1}} = {b_{n + 1}}\]The two conditions are met and so by the Alternating Series Test this series is convergent. We’ll leave it to you to verify this series is also absolutely convergent.

Here’s the limit.

\[L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{n + 3}}{{2\left( {n + 1} \right) + 7}}\,\,\frac{{2n + 7}}{{n + 2}}} \right| = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {n + 3} \right)\left( {2n + 7} \right)}}{{\left( {2n + 9} \right)\left( {n + 2} \right)}} = 1\]Again, the ratio test tells us nothing here. We can however, quickly use the divergence test on this. In fact that probably should have been our first choice on this one anyway.

\[\mathop {\lim }\limits_{n \to \infty } \frac{{n + 2}}{{2n + 7}} = \frac{1}{2} \ne 0\]By the Divergence Test this series is divergent.

So, as we saw in the previous two examples if we get \(L = 1\) from the ratio test the series can be either convergent or divergent.

There is one more thing that we should note about the ratio test before we move onto the next section. The last series was a polynomial divided by a polynomial and we saw that we got \(L = 1\) from the ratio test. This will always happen with rational expression involving only polynomials or polynomials under radicals. So, in the future it isn’t even worth it to try the ratio test on these kinds of problems since we now know that we will get \(L = 1\).

Also, in the second to last example we saw an example of an alternating series in which the positive term was a rational expression involving polynomials and again we will always get \(L = 1\) in these cases.

Let’s close the section out with a proof of the Ratio Test.

#### Proof of Ratio Test

First note that we can assume without loss of generality that the series will start at \(n = 1\) as we’ve done for all our series test proofs.

Let’s start off the proof here by assuming that \(L < 1\) and we’ll need to show that \(\sum {{a_n}} \) is absolutely convergent. To do this let’s first note that because \(L < 1\) there is some number \(r\) such that \(L < r < 1\).

Now, recall that,

\[L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right|\]and because we also have chosen \(r\) such that \(L < r\) there is some \(N\) such that if \(n \ge N\) we will have,

\[\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| < r\hspace{0.5in} \Rightarrow \hspace{0.5in}\left| {{a_{n + 1}}} \right| < r\left| {{a_n}} \right|\]Next, consider the following,

\[\begin{align*}\left| {{a_{N + 1}}} \right| & < r\left| {{a_N}} \right|\\ \left| {{a_{N + 2}}} \right| & < r\left| {{a_{N + 1}}} \right| < {r^2}\left| {{a_N}} \right|\\ \left| {{a_{N + 3}}} \right| & < r\left| {{a_{N + 2}}} \right| < {r^3}\left| {{a_N}} \right|\\ & \hspace{0.5in} \vdots \\ \left| {{a_{N + k}}} \right| & < r\left| {{a_{N + k - 1}}} \right| < {r^k}\left| {{a_N}} \right|\end{align*}\]So, for \(k = 1,2,3, \ldots \) we have \(\left| {{a_{N + k}}} \right| < {r^k}\left| {{a_N}} \right|\). Just why is this important? Well we can now look at the following series.

\[\sum\limits_{k = 0}^\infty {\left| {{a_N}} \right|{r^k}} \]This is a geometric series and because \(0 < r < 1\) we in fact know that it is a convergent series. Also because \(\left| {{a_{N + k}}} \right| < {r^k}\left| {{a_N}} \right|\) by the Comparison test the series

\[\sum\limits_{n = N + 1}^\infty {\left| {{a_n}} \right|} = \sum\limits_{k = 1}^\infty {\left| {{a_{N + k}}} \right|} \]is convergent. However since,

\[\sum\limits_{n = 1}^\infty {\left| {{a_n}} \right|} = \sum\limits_{n = 1}^N {\left| {{a_n}} \right|} + \sum\limits_{n = N + 1}^\infty {\left| {{a_n}} \right|} \]we know that \(\sum\limits_{n = 1}^\infty {\left| {{a_n}} \right|} \) is also convergent since the first term on the right is a finite sum of finite terms and hence finite. Therefore \(\sum\limits_{n = 1}^\infty {{a_n}} \) is absolutely convergent.

Next, we need to assume that \(L > 1\) and we’ll need to show that \(\sum {{a_n}} \) is divergent. Recalling that,

\[L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right|\]and because \(L > 1\) we know that there must be some \(N\) such that if \(n \ge N\) we will have,

\[\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| > 1\hspace{0.5in} \Rightarrow \hspace{0.5in}\left| {{a_{n + 1}}} \right| > \left| {{a_n}} \right|\]However, if \(\left| {{a_{n + 1}}} \right| > \left| {{a_n}} \right|\) for all \(n \ge N\) then we know that,

\[\mathop {\lim }\limits_{n \to \infty } \left| {{a_n}} \right| \ne 0\]because the terms are getting larger and guaranteed to not be negative. This in turn means that,

\[\mathop {\lim }\limits_{n \to \infty } {a_n} \ne 0\]Therefore, by the Divergence Test \(\sum {{a_n}} \) is divergent.

Finally, we need to assume that \(L = 1\) and show that we could get a series that has any of the three possibilities. To do this we just need a series for each case. We’ll leave the details of checking to you but all three of the following series have \(L = 1\) and each one exhibits one of the possibilities.

\[\begin{align*}& \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} & \hspace{0.5in}&{\mbox{absolutely convergent}}\\ & \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{n}} & \hspace{0.5in}&{\mbox{conditionally convergent}}\\ & \sum\limits_{n = 1}^\infty {\frac{1}{n}} & \hspace{0.5in}&{\mbox{divergent}}\end{align*}\]