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Section 4-8 : Alternating Series Test

4. Determine if the following series converges or diverges.

\[\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 6}}n}}{{{n^2} + 9}}} \]

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Start Solution

First, this is (hopefully) clearly an alternating series with,

\[{b_n} = \frac{n}{{{n^2} + 9}}\]

and it should pretty obvious the \({b_n}\) are positive and so we know that we can use the Alternating Series Test on this series.

It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that!

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Let’s first take a look at the limit,

\[\mathop {\lim }\limits_{n \to \infty } {b_n} = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{{n^2} + 9}} = 0\]

So, the limit is zero and so the first condition is met.

Show Step 3

Now let’s take care of the decreasing check. In this case increasing \(n\) will increase both the numerator and denominator and so we can’t just say that clearly the terms are decreasing as we did in the first few problems.

We will have no choice but to do a little Calculus I work for this problem. Here is the function and derivative for that work.

\[f\left( x \right) = \frac{x}{{{x^2} + 9}}\hspace{0.5in}f'\left( x \right) = \frac{{9 - {x^2}}}{{{{\left( {{x^2} + 9} \right)}^2}}}\]

It should be pretty clear that the function will be increasing in \(0 \le x < 3\) and decreasing in \(x > 3\) (the range of \(x\) that corresponds to our range of \(n\)).

So, the \({b_n}\) do not actually form a decreasing sequence but they are decreasing for \(n > 3\) and so we can say that they are eventually decreasing and as discussed in the notes that will be sufficient for us.

Show Step 4

So, both of the conditions in the Alternating Series Test are met and so the series is convergent.