Section 10.8 : Alternating Series Test
2. Determine if the following series converges or diverges.
\[\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 3}}}}{{{n^3} + 4n + 1}}} \]Show All Steps Hide All Steps
Start SolutionFirst, 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!
Show Step 2Let’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 3Now 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 4So, both of the conditions in the Alternating Series Test are met and so the series is convergent.