121616949-math.257 - as large as we desire EXAMPLE 10.20...

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10.2 Series 243 But s n - s n - 1 = ( a 0 + a 1 + a 2 + · · · + a n ) - ( a 0 + a 1 + a 2 + · · · + a n - 1 ) = a n , so as desired lim n →∞ a n = 0. This theorem presents an easy divergence test: if given a series a n the limit lim n →∞ a n does not exist or has a value other than zero, the series diverges. Note well that the converse is not true: If lim n →∞ a n = 0 then the series does not necessarily converge. EXAMPLE 10.19 Show that X n =1 n n + 1 diverges. We compute the limit: lim n →∞ n n + 1 = 1 = 0 . Looking at the first few terms perhaps makes it clear that the series has no chance of converging: 1 2 + 2 3 + 3 4 + 4 5 + · · · will just get larger and larger; indeed, after a bit longer the series starts to look very much like · · · + 1 + 1 + 1 + 1 + · · · , and of course if we add up enough 1’s we can make the sum
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Unformatted text preview: as large as we desire. EXAMPLE 10.20 Show that ∞ X n =1 1 n diverges. Here the theorem does not apply: lim n →∞ 1 /n = 0, so it looks like perhaps the series con-verges. Indeed, if you have the fortitude (or the software) to add up the first 1000 terms you will find that 1000 X n =1 1 n ≈ 7 . 49 , so it might be reasonable to speculate that the series converges to something in the neigh-borhood of 10. But in fact the partial sums do go to infinity; they just get big very, very slowly. Consider the following: 1 + 1 2 + 1 3 + 1 4 > 1 + 1 2 + 1 4 + 1 4 = 1 + 1 2 + 1 2...
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