121616949-math.270 - begins to look as if each term is 1 5...

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256 Chapter 10 Sequences and Series Exercises Determine whether each series converges absolutely, converges conditionally, or diverges. 1. X n =1 ( - 1) n - 1 1 2 n 2 + 3 n + 5 2. X n =1 ( - 1) n - 1 3 n 2 + 4 2 n 2 + 3 n + 5 3. X n =1 ( - 1) n - 1 ln n n 4. X n =1 ( - 1) n - 1 ln n n 3 5. X n =2 ( - 1) n 1 ln n 6. X n =0 ( - 1) n 3 n 2 n + 5 n 7. X n =0 ( - 1) n 3 n 2 n + 3 n 8. X n =1 ( - 1) n - 1 arctan n n Does the series X n =0 n 5 5 n converge? It is possible, but a bit unpleasant, to approach this with the integral test or the comparison test, but there is an easier way. Consider what happens as we move from one term to the next in this series: · · · + n 5 5 n + ( n + 1) 5 5 n +1 + · · · The denominator goes up by a factor of 5, 5 n +1 = 5 · 5 n , but the numerator goes up by much less: ( n + 1) 5 = n 5 + 5 n 4 + 10 n 3 + 10 n 2 + 5 n + 1, which is much less than 5 n 5 when n is large, because 5 n 4 is much less than n 5 . So we might guess that in the long run it
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Unformatted text preview: begins to look as if each term is 1 / 5 of the previous term. We have seen series that behave like this: ∞ X n =0 1 5 n = 5 4 , a geometric series. So we might try comparing the given series to some variation of this geometric series. This is possible, but a bit messy. We can in effect do the same thing, but bypass most of the unpleasant work. The key is to notice that lim n →∞ a n +1 a n = lim n →∞ ( n + 1) 5 5 n +1 5 n n 5 = lim n →∞ ( n + 1) 5 n 5 1 5 = 1 · 1 5 = 1 5 . This is really just what we noticed above, done a bit more officially: in the long run, each term is one fifth of the previous term. Now pick some number between 1 / 5 and 1, say 1 / 2....
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