1/26/2024 0 Comments Diverging sequencesYou may conclude that it cannot be determine because there are restrictions on the direct comparison test (as I just described above). However both of those mistake are fairly common.Ĭannot Be Determined. And even though that is true you may also make the mistake of saying the the given series also diverges, although you cannot compare those two series because the greater diverges, so the smaller could converge or diverge. For example in this problem if you compared the series to 8/n rather than 8/(n^2) you would have come that 8/n diverges. However a common mistake is that when you use any type of comparison test, you must make sure you are comparing a series to the same degree. There are not too many ways to come up with this conclusion. Since 8/n^2 converges so does the given series since it is smallerĭiverge. Since the nth term divergence test does not work, the easiest way to do this is by the Direct Comparison Test:ģ. Another way is that if someone does the nth term divergence test and gets "0", then they may think it diverges, if they mixed it up with another test, such as the p-series test, where when nġ. One way is that if you write it as a partial fraction you get (1/n - 1/ (n+1)), and since 1/n diverges because it is a harmonic series, one may thing the entire expression is harmonic and diverges. There are a couple of ways one can reach this conclusion. You can also test this by just adding the first 5-6 values of the series.ĭiverges. Just by looking at the terms you can tell that since all the values of the series are less than 1, the sum will never reach 3. ∑ ((1/n) – 1/ (n+1)) = 1-.5+.5-.333+.333+.Įvery value after the first value cancels out so it ends up just equaling the first value which is one.ģ.
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