Ratio Test – Definition, Conditions, and Examples on Series

The ratio test is an important method to learn when analyzing different infinite series. It’s one of the first tests used when assessing the convergence or divergence of a given series – especially the Taylor series. The ratio test can also help us in finding the interval and radius of the interval of a power series making it a very important convergence test.

The ratio test utilizes the $\boldsymbol{n}$th and the $\boldsymbol{(n + 1)}$th term of the series. We can determine the divergence or convergence of certain series by taking the ratios of these two terms and evaluating the ratio’s limit as $\boldsymbol{n}$ approaches infinity.

This article will cover all key components we need to understand when working with series and the ratio test. In our discussion, we will:

• Establish the important conditions we need to check before we can apply the ratio test.
• Observe the different types of series that will benefit from this convergence test the most.
• Understand how the ratio test was established.
• Show you how we can apply the ratio test on different series.

Since we will be dealing with different types of series and studying their divergence as well as convergence, refresh what you know of convergent and divergent series. For now, let’s dive right into understanding the definition of the ratio test.

What is the ratio test?

The ratio test is one of the fastest ways for us to determine whether a series is convergent or not because it only needs the $\boldsymbol{n}$th and the $\boldsymbol{(n + 1)}$th terms of the series.

In the past, we’ve learned that knowing $\lim_{n \rightarrow \infty} a_n = 0$ is not enough to conclude that the series we’re working with is convergent. This where convergence tests such as the ratio test come in handy.

Suppose that we’re working with an infinite series, $\sum_{n = 1}^{\infty} a_n$, the limit of its two successive terms’ ratio can be represented as $L$.

\begin{aligned}\lim_{n \rightarrow \infty} \dfrac{a_{ n + 1}}{a_n} &= L\end{aligned}

The ratio test utilizes the value of $\boldsymbol{L}$ to determine whether a given series is divergent or convergent. When $\boldsymbol{|L| < 1}$, the series, $\sum_{n = 1}^{\infty} a_n$, is convergent. When $\boldsymbol{|L| > 1}$, $\sum_{n = 1}^{\infty} a_n$, is divergent. Here’s the downside of using the ratio test: when $\boldsymbol{|L| = 1}$, we won’t be able to conclude anything.

What are the ratio test rules?

Let’s rewrite what was discussed in the previous section to make the ratio test more straightforward. We’ll begin with the series, $\sum_{n =1}^{\infty} a_n$.

\begin{aligned}\sum_{n =1}^{\infty} a_n  &= a_1 + a_2 + a_3 + …+ a_n + a_{n + 1}+…\end{aligned}

We’ll use the $n$th and $(n + 1)$th terms of this series. The ratio test will depend on the value of $ \boldsymbol{\lim_{n \rightarrow \infty} \left| \dfrac{a_{n+1}}{a_n}\right|}$. Let’s say we have $L$ represent this value then there are possible outcomes for $L$. Here’s a table summarizing the possible conclusions for $L$:

The ratio test is most useful the $n$th and $(n+1)$th terms share a lot of common factors or evaluating the limit of their ratios won’t be problematic. Power series such as the Taylor series benefit from the ratio test. This is why you’ll be seeing the ratio test used in different power series in future discussions, so knowing how to apply this test correctly will come in handy later.

If you want to better understand how the ratio test was established, check the proof shown below. Otherwise, feel free to move to the next section to learn how we can use this convergence test.

Understanding the ratio test formula proof

We’ll use the comparison test to prove the ratio test formula and we’ll break the proof into two parts: 1) when $\boldsymbol{L< 1}$ (the series converges) and 2) when $\boldsymbol{L> 1}$  (the series diverges). The goal is for us to compare, $\sum_{n = 1}^{\infty} a_n$, with a convergent geometric series.

For the first part of this proof, we’ll focus on the case when $L <r< 1$.

\begin{aligned}\lim_{n \rightarrow \infty} \left|\dfrac{a_{n + 1}}{a_n}\right| = L, \phantom{x} L < r\end{aligned}

At some point, the ratio, $\left|\dfrac{a_{n+1}}{a_n} \right|$,will be less than $r$. We can let $N$ be an integer that satisfies the inequality shown below.

\begin{aligned} \left| \dfrac{a_{n +1}}{a_n}\right| < r, \phantom{x} n \geq N\end{aligned}

This is actually how we formally define limits, so in case you need a refresher, head over to this link. For now, let’s rewrite the inequality as shown below.

\begin{aligned}|a_{n + 1}| &< |a_n|r, \phantom{x} n \geq N\end{aligned}

We’ll use this inequality to write the inequalities for $N$, $N+1$, $N +2$, and so on as shown below.

\begin{aligned}|a_{N + 1}| &< |a_N|r\\ |a_{N + 2}| &< |a_{N +1}|r < |a_N|r^2\\|a_{N + 3}| &< |a_{N +2}|r < |a_N|r^3\end{aligned}

In general, we have $|a_{N +k}| < |a_N|r^k$, where $k \geq 1$.Hence, we have the series shown below.

\begin{aligned} \sum_{k = 1}^{\infty} |a_n|r^k &= |a_N|r + |a_N|r^2 +|a_N|r^3 +|a_N|r^4 +… \end{aligned}

Remember that we’ve established that $0

\begin{aligned}\sum_{n = N + 1}^{\infty} |a_n| &= \sum_{k = 1}^{\infty} |a_{N +k}| \\&= |a_{N + 1}|+ |a_{N + 2}| + |a_{N + 3}| + …,\end{aligned}

This means that our series, $\sum_{n = 1}^{\infty} a_n$ is convergent for this case. Now, let’s work on the second case: when $L > 1$.

Since we’ve established that $\left|\dfrac{a_{n + 1}}{a_n}\right| < 1$ for this case, we have $|a_{N + 1}| > |a_n|$ for all $n \geq N$. This means that

\begin{aligned}\lim_{n\ \rightarrow \infty} a_n \neq 0\end{aligned}

and through the divergence test, the series $\sum_{n = 1}^{\infty} a_n$} is divergent.

There’s nothing we can conclude when we study what happens if $\lim_{n \rightarrow \infty} \left|\dfrac{a_{n + 1}}{a_n}\right| = 1$. The series can either diverge or converge for this case, so the ratio test is not the most helpful when we have $L =1$.

This proof covers all possible cases, so we’ve just confirmed the validity of the ratio test. Now, let’s see how we can apply what we’ve learned to test the convergence of a series using the ratio test.

How to do the ratio test?

The ratio test is a straightforward process and the guidelines below should help in breaking down the steps for the ratio test:

• Take note of the expressions of the $n$th and $n+1$th terms.
• Find the absolute value of the their ratio, $\dfrac{a_{n + 1}}{a_n}$.
• Now, take the limit of the resulting expression, $\left|\dfrac{a_{n + 1}}{a_n} \right|$, by evaluating the limit when $n \rightarrow \infty$.
• Check if the resulting limit, $L$, is less than, greater than or equal to $1$.
• Use the rules for the ratio test to determine whether the series is convergent or divergent.

Let’s use these steps to confirm that the series, $\sum_{n = 1}^{\infty} \dfrac{2n}{n!}$, is indeed convergent. For this series, we have the following $n$th and $(n +1)$th terms:

\begin{aligned}a_n &= \dfrac{2^n}{n!}\\ a_{n + 1} &= \dfrac{2^{n + 1}}{(n + 1)!} \end{aligned}

Now, find the ratio of $a_{n+1}$ and $a_n$. Make sure that the numerator contains $(n +1)$th term and the denominator has the $n$th term.

\begin{aligned} \dfrac{a_{n +1}}{a_{n}} &= \dfrac{\dfrac{2^{n + 1}}{(n + 1)!} }{\dfrac{2^n}{n!}} \\&= \dfrac{2^{n + 1}}{(n + 1)!}  \cdot \dfrac{n!}{2^n}\end{aligned}

Before we simplify this expression, let’s do a quick refresher on factorials. Recall that we have the following factorial definition:

\begin{aligned} n! &= n(n -1)(n -2)(n -3)(n – 4) …(3)(2)(1)\\0! &= 1\end{aligned}

This also means that $(n+ 1)!$ is simply equal to $(n + 1) \cdot n!$. Let’s use this expression to simplify $\dfrac{a_{n + 1}}{a_n}$.

\begin{aligned} \dfrac{a_{n + 1}}{a_n} &= \dfrac{2^{n + 1}}{(n + 1)!}  \cdot \dfrac{n!}{2^n}\\&= \dfrac{2^n \cdot 2}{(n + 1) \cdot n!} \cdot \dfrac{n!}{2^n}\\&= \dfrac{2}{n + 1}\end{aligned}

Take the absolute value of the expression. Hence, we have $ \left|\dfrac{a_{n + 1}}{a_n} \right| =\dfrac{2}{n + 1}$. Evaluate its limit, $L$, as $n$ approaches $\infty$ to show that the series is indeed convergence.

\begin{aligned} L &= \lim_{n \rightarrow \infty}\left|\dfrac{a_{n + 1}}{a_n} \right|\\&= \lim_{n \rightarrow \infty}\dfrac{2}{n + 1} \\&= \lim_{n \rightarrow \infty} \dfrac{\dfrac{2}{n}}{\dfrac{n}{n} + \dfrac{1}{n}}\\&= \dfrac{0}{1 + 0}\\&= 0 < 1\end{aligned}

Since $L < 1$, through the ratio test, we can confirm that the series, $\sum_{n = 1}^{\infty} \dfrac{2n}{n!}$, is indeed convergent.

Now, it’s your time to use what you’ve just learned, so try working on the problems we’ve prepared for you!  

Example 1

Determine whether the series, $\sum_{n = 1}^{\infty} \dfrac{n^5}{5^n}$ is convergent or divergent.

Solution

Let’s first write down the $n$th and $(n +1)$th terms of the series.

\begin{aligned} a_n &= \dfrac{n^5}{5^n}\\ a_{n + 1} &= \dfrac{(n +1)^5}{5^{n + 1}}\end{aligned}

Now, let’s write the ratio of the two expressions then simplify the expression for $\left|\dfrac{a_{n + 1}}{a_n}\right|$ using exponent rules.

\begin{aligned}\left|\dfrac{a_{n + 1}}{a_n}\right| &= \left| \dfrac{\dfrac{(n +1)^5}{5^{n + 1}}}{\dfrac{n^5}{5^n}}\right| \\&= \left|\dfrac{(n +1)^5}{5^{n} \cdot 5} \cdot \dfrac{5^n}{n^5} \right|\\ &= \left|\dfrac{(n + 1)^5}{5n^5}\right| \\&= \dfrac{1}{5}\left|\dfrac{n^5}{(n + 1)^5}\right|\end{aligned}

Now, evaluate $\left|\dfrac{a_{n + 1}}{a_n} \right|$ as $n \rightarrow \infty$. Using the limit laws, factor out $5$ from the expression then evaluate the limit of the base and raise the result by $5$.

\begin{aligned} L &= \lim_{n \rightarrow \infty}\left|\dfrac{a_{n + 1}}{a_n}\right| \\&= \lim_{n \rightarrow \infty} \dfrac{1}{5} \left|\dfrac{(n + 1)^5}{n^5}\right|\\ &= \dfrac{1}{5} \lim_{n \rightarrow \infty} \left|\dfrac{(n + 1)^5}{n^5}\right|\\&= \dfrac{1}{5} \lim_{n \rightarrow \infty} \left|\left(\dfrac{n  +1}{n} \right )^5\right|\\ &= \dfrac{1}{5} \left| \left(\lim_{n \rightarrow \infty} \dfrac{n + 1}{n} \right )^5\right|\\ &= \dfrac{1}{5}\left|\left(\lim_{n \rightarrow \infty} 1 + \dfrac{1}{n}\right)^5  \right|\\ &= \dfrac{1}{5} \left|\left(1 + 0 \right )^5\right|\\&= \dfrac{1}{5} \end{aligned}

From this, we can see that $ L = \dfrac{1}{5} < 1$, this means that the series, $\sum_{n = 1}^{\infty} \dfrac{n^5}{5^n}$, is a convergent series.

Example 2

Determine whether the series, $\sum_{n = 1}^{\infty} \dfrac{(n + 2)!}{(n – 1)!}$, is convergent or divergent.

Solution

We’ll once again begin by writing the expressions for $a_n$ and $a_{n + 1}$.

\begin{aligned}a_{n} &= \dfrac{2(n + 2)!}{(n – 1)!}\\a_{n + 1} &= \dfrac{[(n + 1) + 2]!}{[(n +1) – 1]!}\\&= \dfrac{(n + 3)!}{n!}\end{aligned}

Now, take the ratio of $a_{n + 1}$ and ${a_n}$ then take the absolute value of the resulting expression. Simplify the rational expression by rewriting $(n + 3)!$ as $(n + 2)! (n + 3)$ and $n!$ as $ (n -1)! n $

\begin{aligned}\left|\dfrac{a_{n + 1}}{a_n}\right|&= \left|\dfrac{(n + 2)!}{(n – 1)!} \cdot  \dfrac{n!}{(n + 3)!}\right| \\&= \left|\dfrac{(n + 2)!}{(n  – 1)!} \cdot  \dfrac{(n -1)!n}{(n + 2)!(n+ 3)}\right|\\ &= \left|\dfrac{\cancel{(n + 2)!}}{\cancel{(n – 1)!}} \cdot  \dfrac{\cancel{(n – 1)!}n}{\cancel{(n + 2)!}(n + 3)}\right|\\ &= \left|\dfrac{n}{n + 3}\right|\end{aligned}

Take the limit of $\left|\dfrac{a^{n + 1}}{a_n}\right|$ as $n$ approaches infinity to test the convergence (or divergence) of the series.

\begin{aligned}L &= \lim_{n \rightarrow \infty}\left|\dfrac{a_{n + 1}}{a_n}\right|\\ &= \lim_{n \rightarrow \infty}\left|\dfrac{n}{n + 3} \right|\\ &= \left|\lim_{n \rightarrow \infty}\dfrac{n}{n + 3}\right|\\&= \left|\lim_{n \rightarrow \infty} \left(\dfrac{1}{1 + \dfrac{3}{n}} \right )\right|\\&= \left|\dfrac{1}{ 1 + 0}\right|\\ &= 1\end{aligned}

Since $\boldsymbol{L = 1}$, we can’t use the ratio test to confirm the convergence of the series. We’ll need to use different tests but before we do so, let’s rule out the possibility that the series diverges using the divergent test.

Take the limit of $a_n$ as $n \rightarrow \infty$ and when the limit is not equal to zero, the series is divergent.

\begin{aligned}\lim_{n \rightarrow \infty} a_n &= \lim_{n \rightarrow \infty} \dfrac{(n + 2)!}{(n – 1)!}\\&= \lim_{n \rightarrow \infty} \dfrac{(n – 1)! \cdot n(n + 1)(n + 2)}{(n -1)!}\\&= \lim_{n \rightarrow \infty} n(n +1) (n +2)\\&= \infty\\ &\neq 0\end{aligned}

Since the limit is not equal to $0$, the series, $\sum_{n = 1}^{\infty} \dfrac{(n + 2)!}{(n – 1)!}$, is divergent.

Example 3

Determine whether the series, $\sum_{n = 3}^{\infty} \dfrac{e^{3n}}{(n – 2)!}$, is convergent or divergent.

Solution

Write the expressions for $a_n$ and $a_{n + 1}$. This might be tricky for this example, so it pays to double-check your work each step of the line.

\begin{aligned}a_n &=\dfrac{e^{3n}}{(n – 2)!}\\ a_{n + 1} &= \dfrac{e^{3(n + 1)}}{[(n + 1) – 2]!}\\&= \dfrac{e^{3n + 3}}{(n -1)!}\end{aligned}

Now, write the expression for the ratio of $a_{n +1}$ and $a_n$ the simplify the result of its absolute value, $\left|\dfrac{a_{n +1}}{a_n}\right|$.

\begin{aligned}\left|\dfrac{a_{n + 1}}{a_n}\right| &=  \left|\dfrac{e^{3n + 3}}{(n -1)!} \cdot \dfrac{(n -2)!}{e^{3n}}\right|\\&= \left|\dfrac{e^{3n} \cdot e^3}{(n – 1)(n -2)!} \cdot \dfrac{(n -2)!}{e^{3n}}\right|\\&= \left|\dfrac{\cancel{e^{3n}} \cdot e^3}{(n – 1)\cancel{(n -2)!}} \cdot \dfrac{\cancel{(n -2)!}}{\cancel{e^{3n}}}\right|\\ &= \left|\dfrac{e^3}{n – 1}\right| \end{aligned}

Now that we have the simplified expression for the $n$th and $(n + 1)$th terms’ ratio, let’s evaluate its limit as $n \rightarrow \infty$.

\begin{aligned}L &= \lim_{n \rightarrow \infty}\left|\dfrac{a_{n + 1}}{a_n}\right|\\ &= \lim_{n \rightarrow \infty}\left|\dfrac{e^3}{n – 1} \right|\\&= \left|\lim_{n \rightarrow \infty}\dfrac{e^3}{n – 1} \right|\\ &= 0\end{aligned}

Since $L = 0 <1$ and through the ratio test, we can conclude that the series, $\sum_{n = 3}^{\infty} \dfrac{e^{3n}}{(n – 2)!}$, is convergent.

Example 4

Determine whether the series, $\sum_{n = 1}^{\infty} \dfrac{n^n}{n!}$, is convergent or divergent. Hint: $\lim_{n \rightarrow \infty} \left(1 + \dfrac{1}{n}\right)^n  = e$.

Solution

Let’s begin by writing down the expressions for the $n$th and the $(n + 1)$th terms.

\begin{aligned}a_n &= \dfrac{n^n}{n!}\\ a_{n + 1} &= \dfrac{(n + 1)^{n + 1}}{(n +1)!}\end{aligned}

Now, take the ratio of these two succeeding terms then take the expression’s absolute value. Simplify the resulting expression when possible.

\begin{aligned}\left|\dfrac{a_{n + 1}}{a_n}\right| &= \left|\dfrac{(n + 1)^{n + 1}}{(n +1)!} \cdot \dfrac{n!}{n^n}\right|\\ &=  \left|\dfrac{(n + 1)\cdot (n + 1)^{n}}{(n +1)\cdot n!} \cdot \dfrac{n!}{n^n}\right|\\ &=\left|\dfrac{\cancel{(n + 1)}\cdot (n + 1)^{n}}{\cancel{(n +1)}\cdot \cancel{n!}} \cdot \dfrac{\cancel{n!}}{n^n}\right|\\ &= \left|\left(\dfrac{n +1}{n} \right )^n\right|\end{aligned}

Now let’s take the limit of the resulting expression knowing that $\lim_{n \rightarrow \infty} \left(1 + \dfrac{1}{n}\right)^n  = e$.

\begin{aligned}L &= \lim_{n \rightarrow \infty}\left|\dfrac{a_{n + 1}}{a_n}\right|\\ &= \lim_{n \rightarrow \infty}\left|\left(1 + \dfrac{1}{n} \right )^n \right|\\&= \left|\lim_{n \rightarrow \infty} \left(1 + \dfrac{1}{n} \right )^n\right|\\ &= e\\ &\approx 2.71828…\end{aligned}

Since $L = e >1$, through the ratio test, we can conclude that the series, $\sum_{n = 1}^{\infty} \dfrac{n^n}{n!}$, is divergent.

Practice Questions

1. Determine whether the series, $\sum_{n = 1}^{\infty} \dfrac{3^n}{n!}$ is convergent or divergent. If you used the ratio test, what was the resulting limit of the ratio?

2. Determine whether the series, $\sum_{n = 1}^{\infty} \dfrac{n^3}{3^n}$ is convergent or divergent. If you used the ratio test, what was the resulting limit of the ratio?

3. Determine whether the series, $\sum_{n = 1}^{\infty} \dfrac{(n + 3)!}{n!}$, is convergent or divergent. If you used the ratio test, what was the resulting limit of the ratio?

4. Determine whether the series, $\sum_{n = 1}^{\infty} \dfrac{n^n }{(n – 2)!}$, is convergent or divergent. If you used the ratio test, what was the resulting limit of the ratio?

5. Determine whether the series, $\sum_{n = 1}^{\infty} \dfrac{e^{4n}}{(n – 2)!}$, is convergent or divergent. If you used the ratio test, what was the resulting limit of the ratio?

6. Determine whether the series, $\sum_{n = 1}^{\infty}(-1)^n \dfrac{(n!)^2 }{(2n)!}$, is convergent or divergent. If you used the ratio test, what was the resulting limit of the ratio?

Answer Key

1. The resulting limit is $0$. Hence, $\sum_{n = 1}^{\infty} \dfrac{3^n}{n!}$ is convergent.

2. The resulting limit is $\dfrac{1}{3}$. Hence, $\sum_{n = 1}^{\infty} \dfrac{n^3}{3^n}$ is convergent.

3. The series, $\sum_{n = 1}^{\infty} \dfrac{(n + 3)!}{(n – 2)!}$, is divergent.

4. The series, $\sum_{n = 1}^{\infty} \dfrac{(n + 3)!}{n!}$, is divergent.

5. The resulting limit is $0 $. Hence, $\sum_{n = 1}^{\infty} \dfrac{e^{4n}}{(n – 2)!}$ is convergent.

6. The resulting limit is $\dfrac{1}{4}$. Hence, $\sum_{n = 1}^{\infty}(-1)^n \dfrac{(n!)^2 }{(2n)!}$ is convergent.