Comparison Test

# Comparison Test - Comparison Test / Limit Comparison Test...

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Comparison Test / Limit Comparison Test In the previous section we saw how to relate a series to an improper integral to determine the convergence of a series. While the integral test is a nice test, it does force us to do improper integrals which aren’t always easy and in some cases may be impossible to determine the convergence of. For instance consider the following series. In order to use the Integral Test we would have to integrate and I’m not even sure if it’s possible to do this integral. Nicely enough for us there is another test that we can use on this series that will be much easier to use. First, let’s note that the series terms are positive. As with the Integral Test that will be important in this section. Next let’s note that we must have since we are integrating on the interval . Likewise, regardless of the value of x we will always have . So, if we drop the x from the denominator the denominator will get smaller and hence the whole fraction will get larger. So, Now, is a geometric series and we know that since the series will converge and its value will be,

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Now, if we go back to our original series and write down the partial sums we get, Since all the terms are positive adding a new term will only make the number larger and so the sequence of partial sums must be an increasing sequence. Then since, and because the terms in these two sequences are positive we can also say that, Therefore, the sequence of partial sums is also a bounded sequence. Then from the second section on sequences we know that a monotonic and bounded sequence is also convergent.
So, the sequence of partial sums of our series is a convergent sequence. This means that the series itself, is also convergent. So, what did we do here? We found a series whose terms were always larger than the original series terms and this new series was also convergent. Then since the original series terms were positive (very important) this meant that the original series was also convergent. To show that a series (with only positive terms) was divergent we could go through a similar argument and find a new divergent series whose terms are always smaller than the original series. In this case the original series would have to take a value larger than the new series. However, since the new series is divergent its value will be infinite. This means that the original series must also be infinite and hence divergent. We can summarize all this in the following test. Comparison Test Suppose that we have two series and with for all n and for all n . Then, 1. If is convergent then so is . 2. If is divergent then so is . In other words, we have two series of positive terms and the terms of one of the series is always larger than the terms of the other series. Then if the larger series is convergent the smaller series must also be convergent. Likewise, if the smaller series is divergent then the larger series must also be divergent. Note as well that in order to apply this test we need both series to start at the same place. A formal

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## This note was uploaded on 11/10/2011 for the course MATH 136 taught by Professor Prellis during the Fall '08 term at Rutgers.

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Comparison Test - Comparison Test / Limit Comparison Test...

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