**Nth Term Test for Divergence**

Today we’re going to work with **limits**, specifically the limit of a series. In order to learn something about the limit of a series, we can run a test on that sequence that creates the series.

But before we get into that, let’s back up a bit and get our vocabulary straight by reviewing how a sequence and a series are different.

A **sequence** is a list of numbers in a particular order, like this:

\({1,3,5,7,9,11,13,…}\)

We can also write a sequence as a formula or rule. For our sequence it looks like this:

\(a_{n}=2n-1\)

The letter \(n\) represents the number of the term in the sequence. So the second number in the sequence can be found by substituting 2 for \(n\) in the formula.

\(a_{2}=2(2)-1=3\)

A **series** is a bit different. A series is the sum of all the numbers in a sequence. So it looks like this:

\({1+3+5+7+9+11+13+…}\)

Notice that instead of commas we have addition signs. Each term also has a partial sum:

Term Number | 1 2 3 4 5 6 7 8 |

Series | 1 + 3 + 5 + 7 + 9 +11 + 13 +15 |

Partial Sum | 1 4 9 16 25 36 49 64 |

The **partial sum** is the sum of all the numbers in the series up until that point. So the partial sum of the third term is \(1+3+5\), which is 9. And if we look carefully at our table, we can see

something very interesting if we look for the patterns. Our series is adding the odd positive integers, and the partial sums also look very familiar. They are the perfect squares!

We can also write a series using math notation:

\(\displaystyle \sum_{n=1}^{\infty }2n-11\)

Look at that fancy symbol! That’s our summation symbol, the Greek letter sigma. And it just means to “sum up” in math. So this expression is asking for the sum of the values of the expression \(2n-1\) for all terms from the first one until infinity. But for our purposes when we see the summation symbol sigma, we know that we are dealing with a *series*. When we see the infinity symbol above the sigma, we know that we are dealing with an **infinite series**. In fact, if we see the sigma without anything above or below it, we can assume that it is an infinite series of the terms from 1 to \(\infty\).

In both science and finance, an infinite series that converges on an actual number is especially useful, so determining whether a series is divergent or convergent is important. In order for a series to be considered convergent, it must pass a sequence of tests. The first test that is used is the \(n^{th}\) **term test**.

The \(n^{th}\) term test can tell us quickly if a series is divergent. It does not tell us if a series is convergent. If a series “passes” the \(n^{th}\) term test, then it must go through a bunch of other tests to be considered convergent. But the \(n^{th}\) term test is very useful to rule out a divergent series so that all those other tests don’t need to be completed.

So what does the \(n^{th}\) term test state? Here it is:

When \(\displaystyle \lim_{ n\to \infty}a_{n}\neq 0\) then \(\displaystyle \sum_{n=1}^{\infty }a_{n}\) is divergent.

This looks complicated but means that when the sequence \(a_{n}\) does not converge on 0, the series won’t converge on anything. In other words, the series is definitely divergent.

Let’s go back to our series of odd numbers that sums to perfect squares and run this test.

When \(\displaystyle \sum_{n=1}^{\infty }2n-1\)

According to the \(n^{th}\) term test, we need to determine if the sequence with the formula \(2n-1\) converges on 0 as \(n\rightarrow \infty\).

The first few terms of our series look like this:

\({1+3+5+7+9+11+13+…}\)

The sequence of that rule looks the same but with commas used instead of addition symbols:

\({1,3,5,7,9,11,13,…}\)

We can easily see that this sequence is not approaching 0. In fact, it’s approaching infinity, as the number will continue to increase by 2 forever. So this sequence is divergent.

That means our series is also divergent because the sequence did not converge on 0.

Already we can see that this test is going to rule out most series that are tested. Let’s test a series that might have a chance.

\(\displaystyle \sum_{n=1}^{\infty }\frac{1}{n}\)

Let’s apply the \(n^{th}\) term test to this series. Let’s quickly review what the \(n^{th}\) term test states:

When \(\displaystyle \lim_{ n\to 0}a_{n}\neq 0\) , then \(\displaystyle \sum_{n=1}^{\infty }a_{n}\) is divergent.

Here our \(a_{n}\) is \(\frac{1}{n}\). So we need to find out if it converges on 0 or not:

Does \(\displaystyle \lim_{ n\to \infty }\frac{1}{n}=0\)?

If we remember how limits of sequences work, we’ll quickly recall that 1n is one of our foundational sequences and does in fact converge on 0. Let’s look at the first several terms:

\(\frac{1}{1},\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6},\frac{1}{7},\frac{1}{8}\)

We can see that the denominator will keep getting bigger and bigger, making our fraction move closer and closer to 0 without ever getting there.

So our series passes the \(n^{th}\) term test! So what does this mean? It means that the infinite series \(\frac{1}{n}\) may or may not be convergent. That’s all the test tells us. To find out if it really is one of those special convergent series, we’d have to run more mathematical tests. Spoiler alert… it isn’t a convergent series. In case you are curious, here are a couple that are convergent:

\(\displaystyle \sum_{n=1}^{\infty }\frac{1}{n^{2}}\text{ and}\sum_{n=1}^{\infty }\frac{1}{2^{n-1}}\)

Can you see why they each passes the \(n^{th}\) term test and are worthy of further consideration? (note: their sequences both converge on 0)

That’s it for the \(n^{th}\) term test for determining whether a series is divergent. I hope this video was helpful. Thanks for watching, and happy studying!