Section 10.8 : Alternating Series Test
3. Determine if the following series converges or diverges.
∞∑n=01(−1)n(2n+3n)Show All Steps Hide All Steps
Start SolutionDo not get excited about the (−1)n is in the denominator! This is still an alternating series! All the (−1)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,
1(−1)n(2n+3n)=(−1)n(−1)n1(−1)n(2n+3n)=(−1)n(−1)2n(2n+3n)=(−1)n(2n+3n)Note that (−1)2n=1 because the exponent is always even!
So, we now know that this is an alternating series with,
bn=12n+3nand it should pretty obvious the bn 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,
limSo, 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}{{{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.
Show Step 4So, both of the conditions in the Alternating Series Test are met and so the series is convergent.