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### Section 10.8 : Alternating Series Test

2. Determine if the following series converges or diverges.

$\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 3}}}}{{{n^3} + 4n + 1}}}$

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

${b_n} = \frac{1}{{{n^3} + 4n + 1}}$

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,

$\mathop {\lim }\limits_{n \to \infty } {b_n} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^3} + 4n + 1}} = 0$

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 it should be pretty clear that,

$\frac{1}{{{n^3} + 4n + 1}} > \frac{1}{{{{\left( {n + 1} \right)}^3} + 4\left( {n + 1} \right) + 1}}$

since increasing $$n$$ will only increase the denominator and hence force the rational expression to be smaller.

Therefore the $${b_n}$$ form a decreasing sequence.

Show Step 4

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