I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.

Paul

August 27, 2020

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### Section 4-8 : Alternating Series Test

The last two tests that we looked at for series convergence have required that all the terms in the series be positive. Of course there are many series out there that have negative terms in them and so we now need to start looking at tests for these kinds of series.

The test that we are going to look into in this section will be a test for alternating series. An **alternating series** is any series, \(\sum {{a_n}} \), for which the series terms can be written in one of the following two forms.

There are many other ways to deal with the alternating sign, but they can all be written as one of the two forms above. For instance,

\[\begin{align*}{\left( { - 1} \right)^{n + 2}} & = {\left( { - 1} \right)^n}{\left( { - 1} \right)^2} = {\left( { - 1} \right)^n}\\ {\left( { - 1} \right)^{n - 1}} & = {\left( { - 1} \right)^{n + 1}}{\left( { - 1} \right)^{ - 2}} = {\left( { - 1} \right)^{n + 1}}\end{align*}\]There are of course many others, but they all follow the same basic pattern of reducing to one of the first two forms given. If you should happen to run into a different form than the first two, don’t worry about converting it to one of those forms, just be aware that it can be and so the test from this section can be used.

#### Alternating Series Test

Suppose that we have a series \(\sum {{a_n}} \) and either \({a_n} = {\left( { - 1} \right)^n}{b_n}\) or \({a_n} = {\left( { - 1} \right)^{n + 1}}{b_n}\) where \({b_n} \ge 0\) for all \(n\). Then if,

- \(\mathop {\lim }\limits_{n \to \infty } {b_n} = 0\) and,
- \(\left\{ {{b_n}} \right\}\) is a decreasing sequence

the series \(\sum {{a_n}} \) is convergent.

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

There are a couple of things to note about this test. First, unlike the Integral Test and the Comparison/Limit Comparison Test, this test will only tell us when a series converges and not if a series will diverge.

Secondly, in the second condition all that we need to require is that the series terms, \({b_n}\) will be eventually decreasing. It is possible for the first few terms of a series to increase and still have the test be valid. All that is required is that eventually we will have \({b_n} \ge {b_{n + 1}}\) for all \(n\) after some point.

To see why this is consider the following series,

\[\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}{b_n}} \]Let’s suppose that for \(1 \le n \le N\) \(\left\{ {{b_n}} \right\}\) is not decreasing and that for \(n \ge N + 1\) \(\left\{ {{b_n}} \right\}\) is decreasing. The series can then be written as,

\[\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}{b_n}} = \sum\limits_{n = 1}^N {{{\left( { - 1} \right)}^n}{b_n}} + \sum\limits_{n = N + 1}^\infty {{{\left( { - 1} \right)}^n}{b_n}} \]The first series is a finite sum (no matter how large \(N\) is) of finite terms and so we can compute its value and it will be finite. The convergence of the series will depend solely on the convergence of the second (infinite) series. If the second series has a finite value then the sum of two finite values is also finite and so the original series will converge to a finite value. On the other hand, if the second series is divergent either because its value is infinite or it doesn’t have a value then adding a finite number onto this will not change that fact and so the original series will be divergent.

The point of all this is that we don’t need to require that the series terms be decreasing for all \(n\). We only need to require that the series terms will eventually be decreasing since we can always strip out the first few terms that aren’t actually decreasing and look only at the terms that are actually decreasing.

Note that, in practice, we don’t actually strip out the terms that aren’t decreasing. All we do is check that eventually the series terms are decreasing and then apply the test.

Let’s work a couple of examples.

First, identify the \({b_n}\) for the test.

\[\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{n}} = \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n + 1}}\frac{1}{n}} \hspace{0.5in}{b_n} = \frac{1}{n}\]Now, all that we need to do is run through the two conditions in the test.

\[\mathop {\lim }\limits_{n \to \infty } {b_{n}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n} = 0\] \[{b_n} = \frac{1}{n} > \frac{1}{{n + 1}} = {b_{n + 1}}\]Both conditions are met and so by the Alternating Series Test the series must converge.

The series from the previous example is sometimes called the **Alternating Harmonic Series**. Also, the \({\left( { - 1} \right)^{n + 1}}\) could be \({\left( { - 1} \right)^n}\) or any other form of alternating sign and we’d still call it an Alternating Harmonic Series.

In the previous example it was easy to see that the series terms decreased since increasing \(n\) only increased the denominator for the term and hence made the term smaller. In general however, we will need to resort to Calculus I techniques to prove the series terms decrease. We’ll see an example of this in a bit.

First, identify the \({b_n}\) for the test.

\[\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}{n^2}}}{{{n^2} + 5}}} = \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\frac{{{n^2}}}{{{n^2} + 5}}} \hspace{0.5in} \Rightarrow \hspace{0.5in}{b_n} = \frac{{{n^2}}}{{{n^2} + 5}}\]Let’s check the conditions.

\[\mathop {\lim }\limits_{n \to \infty } {b_{n}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^2}}}{{{n^2} + 5}} = 1 \ne 0\]So, the first condition isn’t met and so there is no reason to check the second. Since this condition isn’t met we’ll need to use another test to check convergence. In these cases where the first condition isn’t met it is usually best to use the divergence test.

So, the divergence test requires us to compute the following limit.

\[\mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( { - 1} \right)}^n}{n^2}}}{{{n^2} + 5}}\]This limit can be somewhat tricky to evaluate. For a second let’s consider the following,

\[\mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( { - 1} \right)}^n}{n^2}}}{{{n^2} + 5}} = \left( {\mathop {\lim }\limits_{n \to \infty } {{\left( { - 1} \right)}^n}} \right)\left( {\mathop {\lim }\limits_{n \to \infty } \frac{{{n^2}}}{{{n^2} + 5}}} \right)\]Splitting this limit like this can’t be done because this operation requires that both limits exist and while the second one does the first clearly does not. However, it does show us how we can at least convince ourselves that the overall limit does not exist (even if it won’t be a direct proof of that fact).

So, let’s start with,

\[\mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( { - 1} \right)}^n}{n^2}}}{{{n^2} + 5}} = \mathop {\lim }\limits_{n \to \infty } \left[ {{{\left( { - 1} \right)}^n}\frac{{{n^2}}}{{{n^2} + 5}}} \right]\]Now, the second part of this clearly is going to 1 as \(n \to \infty \) while the first part just alternates between 1 and -1. So, as \(n \to \infty \) the terms are alternating between positive and negative values that are getting closer and closer to 1 and -1 respectively.

In order for limits to exist we know that the terms need to settle down to a single number and since these clearly don’t this limit doesn’t exist and so by the Divergence Test this series diverges.

Notice that in this case the exponent on the “-1” isn’t \(n\) or \(n + 1\). That won’t change how the test works however so we won’t worry about that. In this case we have,

\[{b_n} = \frac{{\sqrt n }}{{n + 4}}\]so let’s check the conditions.

The first is easy enough to check.

\[\mathop {\lim }\limits_{n \to \infty } {b_n} = \mathop {\lim }\limits_{n \to \infty } \frac{{\sqrt n }}{{n + 4}} = 0\]The second condition requires some work however. It is not immediately clear that these terms will decrease. Increasing \(n\) to \(n + 1\) will increase both the numerator and the denominator. Increasing the numerator says the term should also increase while increasing the denominator says that the term should decrease. Since it’s not clear which of these will win out we will need to resort to Calculus I techniques to show that the terms decrease.

Let’s start with the following function and its derivative.

\[f\left( x \right) = \frac{{\sqrt x }}{{x + 4}}\hspace{0.5in}\hspace{0.25in}f'\left( x \right) = \frac{{4 - x}}{{2\sqrt x {{\left( {x + 4} \right)}^2}}}\]Now, there are two critical points for this function, \(x = 0\), and \(x = 4\). Note that \(x = - 4\) is not a critical point because the function is not defined at \(x = - 4\). The first is outside the bound of our series so we won’t need to worry about that one. Using the test points,

\[f'\left( 1 \right) = \frac{3}{{50}}\hspace{0.5in}f'\left( 5 \right) = - \frac{{\sqrt 5 }}{{810}}\]and so we can see that the function in increasing on \(0 \le x \le 4\) and decreasing on \(x \ge 4\). Therefore, since \(f\left( n \right) = {b_n}\) we know as well that the \({b_n}\) are also increasing on \(0 \le n \le 4\) and decreasing on \(n \ge 4\).

The \({b_n}\) are then eventually decreasing and so the second condition is met.

Both conditions are met and so by the Alternating Series Test the series must be converging.

As the previous example has shown, we sometimes need to do a fair amount of work to show that the terms are decreasing. Do not just make the assumption that the terms will be decreasing and let it go at that.

Let’s do one more example just to make a point.

The point of this problem is really just to acknowledge that it is in fact an alternating series. To see this we need to acknowledge that,

\[\cos \left( {n\pi } \right) = {\left( { - 1} \right)^n}\]If you aren’t sure of this you can easily convince yourself that this is correct by plugging in a few values of \(n\) and checking.

So the series is really,

\[\sum\limits_{n = 2}^\infty {\frac{{\cos \left( {n\pi } \right)}}{{\sqrt n }}} = \sum\limits_{n = 2}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{\sqrt n }}\hspace{0.5in} \Rightarrow \hspace{0.25in}\,\,\,\,\,{b_n} = \frac{1}{{\sqrt n }}} \]Checking the two condition gives,

\[\mathop {\lim }\limits_{n \to \infty } {b_n} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{\sqrt n }} = 0\] \[{b_n} = \frac{1}{{\sqrt n }} > \frac{1}{{\sqrt {n + 1} }} = {b_{n + 1}}\]The two conditions of the test are met and so by the Alternating Series Test the series is convergent.

It should be pointed out that the rewrite we did in previous example only works because \(n\) is an integer and because of the presence of the \(\pi\). Without the \(\pi\) we couldn’t do this and if \(n\) wasn’t guaranteed to be an integer we couldn’t do this.

Let’s close this section out with a proof of the Alternating Series Test.

#### Proof of Alternating Series Test

Without loss of generality we can assume that the series starts at \(n = 1\). If not we could modify the proof below to meet the new starting place or we could do an index shift to get the series to start at \(n = 1\).

Also note that the assumption here is that we have \({a_n} = {\left( { - 1} \right)^{n+1}}{b_n}\). To get the proof for \({a_n} = {\left( { - 1} \right)^{n}}{b_n}\) we only need to make minor modifications of the proof and so will not give that proof.

Finally, in the examples all we really needed was for the \({b_n}\) to be positive and decreasing eventually but for this proof to work we really do need them to be positive and decreasing for all \(n\).

First, notice that because the terms of the sequence are decreasing for any two successive terms we can say,

\[b_{n} - {b_{n + 1}} \ge 0\]Now, let’s take a look at the even partial sums.

\[\begin{align*}{s_2}& = {b_1} - {b_2} \ge 0\\ {s_4} & = {b_1} - {b_2} + {b_3} - {b_4} = {s_2} + {b_3} - {b_4} \ge {s_2} & \hspace{0.5in} & {\mbox{because }}{b_3} - {b_4} \ge 0\\ {s_6} & = {s_4} + {b_5} - {b_6} \ge {s_4} & \hspace{0.5in} & {\mbox{because }}{b_5} - {b_6} \ge 0\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \vdots \\ {s_{2n}} & = {s_{2n - 2}} + {b_{2n - 1}} - {b_{2n}} \ge {s_{2n - 2}} & \hspace{0.5in} & {\mbox{because }}{b_{2n - 1}} - {b_{2n}} \ge 0\end{align*}\]So, \(\left\{ {{s_{2n}}} \right\}\) is an increasing sequence.

Next, we can also write the general term as,

\[\begin{align*}{s_{2n}} & = {b_1} - {b_2} + {b_3} - {b_4} + {b_5} + \cdots - {b_{2n - 2}} + {b_{2n - 1}} - {b_{2n}}\\ & = {b_1} - \left( {{b_2} - {b_3}} \right) - \left( {{b_4} - {b_5}} \right) + \cdots - \left( {{b_{2n - 2}} - {b_{2n - 1}}} \right) - {b_{2n}}\end{align*}\]Each of the quantities in parenthesis are positive and by assumption we know that \({b_{2n}}\) is also positive. So, this tells us that \({s_{2n}} \le {b_1}\) for all \(n\).

We now know that \(\left\{ {{s_{2n}}} \right\}\) is an increasing sequence that is bounded above and so we know that it must also converge. So, let’s assume that its limit is \(s\) or,

\[\mathop {\lim }\limits_{n \to \infty } {s_{2n}} = s\]Next, we can quickly determine the limit of the sequence of odd partial sums, \(\left\{ {{s_{2n + 1}}} \right\}\), as follows,

\[\mathop {\lim }\limits_{n \to \infty } {s_{2n + 1}} = \mathop {\lim }\limits_{n \to \infty } \left( {{s_{2n}} + {b_{2n + 1}}} \right) = \mathop {\lim }\limits_{n \to \infty } {s_{2n}} + \mathop {\lim }\limits_{n \to \infty } {b_{2n + 1}} = s + 0 = s\]So, we now know that both \(\left\{ {{s_{2n}}} \right\}\) and \(\left\{ {{s_{2n + 1}}} \right\}\) are convergent sequences and they both have the same limit and so we also know that \(\left\{ {{s_n}} \right\}\) is a convergent sequence with a limit of \(s\). This in turn tells us that \(\sum {{a_n}} \) is convergent.