Let’s Get Real… Analysis (Part 6): The Ratio Test for Sequences

Once you learn how to use the definition of a sequence limit to prove that a sequence $(a_n)$ converges to a limit, the next step is to develop a set of tools that you can use so that way you can solve more challenging problems, without having to do a proof by definition every time. Some of these tools are the Squeeze Theorem and the Monotone Convergence Theorem. Today, we will add another tool to our tool kit: the ratio test.

The Ratio Test for Sequences

The Ratio Test for Sequences: Let $(a_n)$ be a sequence where $a_n \neq 0$ for all $n$ and whose ratio has the limit,

$A = \lim_{n\rightarrow \infty} \frac{|a_{n+1}|}{|a_n|}.$

Then, we have the following implications.

If $A<1,$ then $a_n \rightarrow 0.$
If $A>1,$ then $a_n$ diverges (unbounded).
If $A=1,$ then ???????????????????????????????????????????????????????????

Proof Prolegomenon: Vocabulary aside, prolegomenon is like a prologue to a book, but it is used in more scholarly work in order to set the stage for more detailed analysis later on. And I suppose that this is somewhat scholarly, but most importantly, it’s a really fun word to say! Try it! Basically, we are going to give an outline for the following proof.

We have three parts to prove! How wonderful is that? For part 1, the key idea is to rely on the squeeze theorem to prove $|a_n|\rightarrow 0.$ But, in order to do so, we must find another two sequences to sandwich $|a_n|$ between! Our lower bound sequence will be the constant sequence $(0) = 0,0,0,\cdots,0 \rightarrow 0$ and our upper bound sequence will be a geometric sequence made from some $A<x<1.$ We will show (Lemma 1) that choosing $A<x<1$ leads to $C\cdot x^n\rightarrow 0$ which for a certain value of $C$ will be our geometric sequence. These together prove that $|a_n|\rightarrow 0.$

Lemma 1: Let $0\leq x<1$ and $C \in \mathbb{R}.$ Then, the geometric sequence $C\cdot x^n$ converges to zero.

Proof: We will use the monotone convergence theorem (love this theorem). If it happens to be the case that $x=0,$ then $C\cdot x^n = 0$ for all $n.$ Which makes it clear that $C\cdot x^n\rightarrow 0,$ likewise when $C=0.$ For these reasons, assume $0<x<1$ and consider the case where $C>0.$ Observe that $C\cdot x^n$ is monotone decreasing and bounded below. These follow from the fact that $0< x<1$ implies $x^{n+1}< x^n$ and the assumption that both $x$ and $C$ are non-negative imply that $C\cdot x^n,$ is bounded below by 0. Therefore, by the monotone convergence theorem we deduce the following,

$Cx^n \rightarrow CX$ ,
$Cx^{n+1} \rightarrow CX$ ,
$Cx^{n+1} = Cx^n x\rightarrow CX x$ ,

Where we used some limit laws that are intuitive, but we have yet to prove. We will prove these next time, but until then you can try to prove them yourself! Using the last two bullet points we end up with the algebraic equation $CX x = CX,$ which becomes $X(x-1) = 0.$ Since we assumed $x<1$ it must be that $X=0.$

The case where $C<0$ is similar. Give it a go! You can do it!

$\square_{\mathrm{Lemma}\;1}$

We are not done yet; we will need one more result. We want to be able to use $|a_n|\rightarrow 0$ to conclude that $a_n\rightarrow 0.$ Better yet, we can prove the stronger statement given in Lemma 2.

Lemma 2: $\;\;\textit{ }\;\;a_n\rightarrow 0 \iff |a_n|\rightarrow 0.$

Proof:

Forward: $a_n\rightarrow 0 \iff |a_n|\rightarrow 0.$

Let $a_n\rightarrow 0$ and $\varepsilon >0.$ By the definition of a limit, we know there is some $N\in\mathbb{N}$ such that,

$|a_n - 0| = |a_n| < \varepsilon,$

for all $n>N.$ Using this same $N,$ we have the similar inequality,

$||a_n| -0| = |a_n| < \varepsilon.$

proving $|a_n|\rightarrow 0.$

Backward: $|a_n|\rightarrow 0 \iff a_n\rightarrow 0.$

Give it a go!

$\square_{\mathrm{Lemma}\;2}$

In part 2, we will use a very similar argument in order to compare $|a_n|$ to a divergent geometric series, $x^n,$ when $x>1.$

Part 3, stary tuned!

Proof of Ratio Test:

(1) Let $A<1.$ Can you see why it must be the case that $0\leq A < 1?$ Choose some $x \in (A,1)$ and let $\varepsilon = x - A>0.$ Since $A = \lim_{n\rightarrow \infty} \frac{|a_{n+1}|}{|a_n|},$ by the definition of a limit we know that there is some $N\in\mathbb{N}$ such that,

$\Bigg| \frac{|a_{n+1}|}{|a_n|} - A\Bigg|< \varepsilon =x - A,$

for all $n>N$ . Or, better yet,

$\frac{|a_{n+1}|}{|a_n|} - A<x-A\;$ and therefore $\;\frac{|a_{n+1}|}{|a_n|}<x.$

Now we will employ the trick were we multiply by a fancy looking 1. In this case, we will multiply $|a_n|$ by $\frac{|a_{N}|}{|a_N|} \cdot \frac{|a_{N+1}|}{|a_{N+1}|}\cdot \frac{|a_{N+2}|}{|a_{N+2}|} \cdot \frac{|a_{N+3}|}{|a_{N+3}|} \cdots \frac{|a_{n-1}|}{|a_{n-1}|},$ where we have $n>N$ . This gives,

$|a_n| = |a_N| \cdot \frac{|a_{N+1}|}{|a_{N}|}\cdot \frac{|a_{N+2}|}{|a_{N+1}|} \cdot \frac{|a_{N+3}|}{|a_{N+2}|} \cdots \frac{|a_{n}|}{|a_{n-1}|}.$

Since we are assuming $n>N$ , each fraction is less than $x$ . I.e., $\frac{|a_{N+1}|}{|a_{N+1}|}\,,\, \frac{|a_{N+2}|}{|a_{N+2}|} \,,\,\frac{|a_{N+3}|}{|a_{N+3}|}\,,\,\cdots\,,\, \frac{|a_{n}|}{|a_{n-1}|}<x.$ Thus,

$|a_n| < |a_N| \cdot x\cdot x\cdots x =|a_N| \cdot x^{n-N}.$

We can now use the squeeze theorem, because $0<|a_n| <\frac{|a_N|}{x^N} \cdot x^{n}$ (for all $n>N$ ) and by lemma 1 the sequence $\frac{|a_N|}{x^N} \cdot x^{n} \rightarrow 0.$ Thus, by the squeeze theorem (nuanced) (see footnote),¹

$|a_n| \rightarrow 0.$

Finally, by Lemma 2, $|a_n| \rightarrow 0$ is equivalent to $a_n \rightarrow 0.$ Just like that, we’re done… part 1.

(2) Let $A>1.$ We aim to show that $a_n$ diverges and we do so by proving that $a_n$ is unbounded. With this goal in mind, we will use a similar argument to the one used in part 1. First, note there is some $x$ such that $A>x>1$ and furthermore let $\varepsilon = A-x>0$ there is some $N\in\mathbb{N}$ such that,

$\Bigg| \frac{|a_{n+1}|}{|a_n|} - A\Bigg|< \varepsilon = A-x,$

for all $n>N$ . Or, better yet,

$-( A-x)<\frac{|a_{n+1}|}{|a_n|} - A<( A-x).$

therefore $\frac{|a_{n+1}|}{|a_n|} >x$ for all $n>N$ . Again, multiply by a fancy looking 1 we get,

$|a_n| = |a_N| \cdot \frac{|a_{N+1}|}{|a_{N}|}\cdot \frac{|a_{N+2}|}{|a_{N+1}|} \cdot \frac{|a_{N+3}|}{|a_{N+2}|} \cdots \frac{|a_{n}|}{|a_{n-1}|}.$

But this time we end up with,

$|a_n| > |a_N| \cdot x\cdot x\cdots x =|a_N| \cdot x^{n-N}=\frac{|a_N|}{x^N} \cdot x^{n}.$

And since $x^n\rightarrow \infty,$ (I leave that for you to prove, you’re welcome) we conclude $|a_n|$ is unbounded and hence $a_n$ diverges.

(3) Take a look at the sequences: $\frac{1}{n}$ and $(-1)^n.$ They both have $A=1,$ but the former converges and the later doesn’t. This means that the ratio test is absolutely useless when $A=1.$

$\square_{\mathrm{Ratio}\;\mathrm{Test}}$

Let’s Use the Ratio Test

We will end with a quick application of the ratio test. We will prove the following limit:

$n^{1/n}\rightarrow 1.$

Proof: Ironically, we will be using the root test and the definition of the limit. To this end, let $\varepsilon >0$ and consider the sequence $a_n = \frac{n}{(1+\varepsilon)^n }.$ Using the ratio test,

$\frac{\Big|\frac{n+1}{(1+\varepsilon)^{n+1} }\Big|}{\Big|\frac{n}{(1+\varepsilon)^n }\Big|} = \frac{n+1}{n(1+\varepsilon)} = \Big( 1 + \frac{1}{n}\Big)\frac{1}{(1+\varepsilon)} \longrightarrow \Big( 1 + 0\Big)\frac{1}{(1+\varepsilon)} = \frac{1}{(1+\varepsilon)}<1.$

Thus, $a_n = \frac{n}{(1+\varepsilon)^n } \longrightarrow 0.$ Therefore, there exists a $N\in\mathbb{N}$ such that,

$\Big|\frac{n}{(1+\varepsilon)^n } -0\Big|<1 .$

for all $n>N.$ Hence,

$n<(1+\varepsilon)^n \implies n^{1/n}< 1+\varepsilon \implies |n^{1/n} - 1|<\varepsilon,$

for all $n>N.$ Which, by the defininton of limit implies $n^{1/n}\rightarrow 1.$

$\square$

Concluding Remarks

The ratio test is a very useful tool in analysis. We will see it again when we start to work with infinite series. But until then let’s get some practice with the ratio test. See what you can say regarding the following sequences: