Calculus II - Alternating Series Test

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Section 10.8 : Alternating Series Test

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3. Determine if the following series converges or diverges.

\[\sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( { - 1} \right)}^n}\left( {{2^n} + {3^n}} \right)}}} \]

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Start Solution

Do not get excited about the \({\left( { - 1} \right)^n}\) is in the denominator! This is still an alternating series! All the \({\left( { - 1} \right)^n}\) does is change the sign regardless of whether or not it is in the numerator.

Also note that we could just as easily rewrite the terms as,

\[\frac{1}{{{{\left( { - 1} \right)}^n}\left( {{2^n} + {3^n}} \right)}} = \frac{{{{\left( { - 1} \right)}^n}}}{{{{\left( { - 1} \right)}^n}}}\frac{1}{{{{\left( { - 1} \right)}^n}\left( {{2^n} + {3^n}} \right)}} = \frac{{{{\left( { - 1} \right)}^n}}}{{{{\left( { - 1} \right)}^{2n}}\left( {{2^n} + {3^n}} \right)}} = \frac{{{{\left( { - 1} \right)}^n}}}{{\left( {{2^n} + {3^n}} \right)}}\]

Note that \({\left( { - 1} \right)^{2n}} = 1\) because the exponent is always even!

So, we now know that this is an alternating series with,

\[{b_n} = \frac{1}{{{2^n} + {3^n}}}\]

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,

\[\mathop {\lim }\limits_{n \to \infty } {b_n} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{2^n} + {3^n}}} = 0\]

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,

\[\frac{1}{{{2^n} + {3^n}}} > \frac{1}{{{2^{n + 1}} + {3^{n + 1}}}}\]

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.

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