Calculus/Ratio Test

Ratio Test edit

The Ratio Test is probably the most important test and the test you will use the most as you are learning infinite series. It is used A LOT in power series. I believe it is the most powerful test of all. So I suggest you master it from the start. It's not hard, and if your algebra skills are strong, you might even find it fun to use. Also, the more familiar you are with it and the more practice problems you work, the sooner you will start to be able to look at a series and see almost right away if the Ratio Test will tell you what you need to know.

Ratio Test Definition edit

Let $\textstyle {\sum {a_{n}}}$ be a series with nonzero terms and let $\displaystyle {\lim _{n\to \infty }{\left|{\frac {a_{n+1}}{a_{n}}}\right|}=L}$

Three cases are possible depending on the value of L.
$L<1$ : The series converges absolutely.
$L=1$ : The Ratio Test is inconclusive.
$L>1$ : The series diverges.

Ratio Test Quick Notes edit

used to prove convergence	yes
used to prove divergence	yes
can be inconclusive	yes

if $L=\infty$ then $L>1$ and the series diverges
$a_{n}$ terms can be positive or negative or both
requires the calculation of limits at infinity

When To Use the Ratio Test edit

The ratio test is best used when you have certain elements in the sum. The way to get a feel for this is to build a set of tables containing examples of tests that work as you are working practice problems. This is an extremely powerful technique that will help you really understand infinite series.
Here is a list of things to watch for.

Sums that include factorials.
Sums with exponents containing $n$ .

How To Use The Ratio Test edit

In general, the idea is to set up the ratio $\displaystyle {\lim _{n\to \infty }{\left|{\frac {a_{n+1}}{a_{n}}}\right|}=L}$ and evaluate it.
In detail, you need to determine what $a_{n}$ is and then build $a_{n+1}$ , set up the fraction, combine like terms and then take the limit of each term. Setting up the limit and combining like terms are the easy parts. The challenge comes in taking the limit.

  Key - It is important to remember to use the absolute value signs unless you are absolutely convinced that the term will always be positive.  This is critical to practice up front since, once you get to Taylor Series, you can't and don't want to drop the absolute value signs.  They are critical to the result.  It is never wrong to include them and, as you work more problems, you will get a feel for when you need them and when you don't.  In the practice problems and examples, we will use them unless we explicitly state that they are not needed.  Some instructors are less rigid about this than others.  As always, check with your instructor to see what they require.

Things To Notice edit

If you get $L=\infty$ for the limit, this indicates divergence since it fits the case where the limit is greater than one. Notice that the theorem says nothing about the limit needing to be finite.
In the fraction that you are taking the limit of, the $n+1$ term is in the numerator and the $n$ term is in the denominator. In order for the ratio test to work, they must appear exactly like this. Why? Because if the order was reversed, the conclusion would be incorrect. For example, if we write it correctly and we get L=1/2, we know that the series converges (correctly). However, if we switched the numerator and denominator, we get L=2, which indicates the series diverges (incorrectly).
If you get $L=1$ , you cannot say anything about convergence or divergence of the series. You need to use another test. Sometimes, a comparison test (either direct or limit) will be the best next step.

Ratio Test Proof edit

Here are a couple of proofs of the Ratio Test.
This first video contains a rather long and involved proof of the ratio test. It uses a comparison test. Proof of the Ratio Test

Here is a second proof presented in 3 separate videos.

The Ratio Test - Proof of Part (a)
The Ratio Test - Proof of Part (b)
The Ratio Test - Proof of Part (c)

You do not need to watch these proofs in order to use and understand the Ratio Test. We are including them here for those who are interested.

Video Recommendations edit

This first video clip [1min-50secs] is a great overview of the ratio test. Notice that he doesn't use absolute value signs, so he requires that the terms be positive.
Series, Comparison + Ratio Tests

The beginning [9min-19secs] of this next video has a good discussion about the ratio test. Then the instructor shows two examples when the ratio test is inconclusive to emphasize that a series may converge or diverge when the ratio test is inconclusive.
Ratio Test for Convergence

Before You Start Working The Practice Problems edit

Take a few minutes and scan your list of practice problems. You will notice there are a LOT of very different ones. The key to solving infinite series problems is to find patterns so that you can quickly narrow down the techniques that might work to about 2 or 3. This is especially true with problems on which the ratio test works.

Ratio Test Practice Problems edit

Determine the convergence or divergence of these series using the ratio test, if possible. [These instructions imply that if the ratio test fails $(L=1)$ , you need to use another test to prove convergence or divergence.]

Practice Problems with Written Solutions edit

1. $\sum _{n=0}^{\infty }{\left[{\frac {n!}{2^{n}}}\right]}$

Answer

The series diverges by the Ratio Test.

Solution

Whenever you have a factorial, the ratio test will often work. So, let's try it.

\displaystyle {a_{n}={\frac {n!}{2^{n}}}}

and

\displaystyle {a_{n+1}={\frac {(n+1)!}{2^{n+1}}}}

\displaystyle {\lim _{n\to \infty }{\left|{\frac {a_{n+1}}{a_{n}}}\right|}=\lim _{n\to \infty }{\left|{\frac {(n+1)!}{2^{n+1}}}\div {\frac {n!}{2^{n}}}\right|}=\lim _{n\to \infty }{\left|{\frac {(n+1)!}{2^{n+1}}}\cdot {\frac {2^{n}}{n!}}\right|}}

Now combine like terms.

\displaystyle {{\frac {2^{n}}{2^{n+1}}}={\frac {2^{n}}{2(2^{n})}}=1/2}

\displaystyle {{\frac {(n+1)!}{n!}}={\frac {1\cdot 2\cdot 3\cdot 4...n\cdot (n+1)}{1\cdot 2\cdot 3\cdot 4...n}}=n+1}

So our limit is now

\displaystyle {\lim _{n\to \infty }{\left|{\frac {n+1}{2}}\right|}={\frac {\infty }{2}}=\infty >1\to }

the series diverges.
Note - We could have left off the absolute value signs all the way through this problem since the terms are always positive. However, to follow the theorem exactly, we need them unless we explicitly state that the ratio is positive. It is a common practice (although not always good) to leave them off without explaining, which teachers may do quite often. If you don't understand something, always ask.

Whenever you have a factorial, the ratio test will often work. So, let's try it.

\displaystyle {a_{n}={\frac {n!}{2^{n}}}}

and

\displaystyle {a_{n+1}={\frac {(n+1)!}{2^{n+1}}}}

\displaystyle {\lim _{n\to \infty }{\left|{\frac {a_{n+1}}{a_{n}}}\right|}=\lim _{n\to \infty }{\left|{\frac {(n+1)!}{2^{n+1}}}\div {\frac {n!}{2^{n}}}\right|}=\lim _{n\to \infty }{\left|{\frac {(n+1)!}{2^{n+1}}}\cdot {\frac {2^{n}}{n!}}\right|}}

Now combine like terms.

\displaystyle {{\frac {2^{n}}{2^{n+1}}}={\frac {2^{n}}{2(2^{n})}}=1/2}

\displaystyle {{\frac {(n+1)!}{n!}}={\frac {1\cdot 2\cdot 3\cdot 4...n\cdot (n+1)}{1\cdot 2\cdot 3\cdot 4...n}}=n+1}

So our limit is now

\displaystyle {\lim _{n\to \infty }{\left|{\frac {n+1}{2}}\right|}={\frac {\infty }{2}}=\infty >1\to }

the series diverges.
Note - We could have left off the absolute value signs all the way through this problem since the terms are always positive. However, to follow the theorem exactly, we need them unless we explicitly state that the ratio is positive. It is a common practice (although not always good) to leave them off without explaining, which teachers may do quite often. If you don't understand something, always ask.

Practice Problems with Video Solutions edit

1	$\sum _{k=1}^{\infty }{\left[{\frac {3^{k}}{k^{2}}}\right]}$	answer diverges diverges	solution		2	$\sum _{n=1}^{\infty }{\left[{\frac {(2n)!}{n!n!}}\right]}$	answer diverges diverges	solution