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**Unformatted text preview: **Tests for convergence I Now we’re focused on finding out when an infinite series ∞ X k =0 a k converges or diverges. So far we know only about one kind of infinite series, the geometric series ∞ X k =0 r k . This converges if | r | < 1 and diverges if | r | ≥ 1. We even know what it adds up to (see previous notes). The only other type of infinite series where, when it converges, we know what it adds up to is called a telescoping series . A telescoping series is of the form ∞ X k =0 ( a k +1- a k ) or ∞ X k =0 ( a k- a k +1 ) Notice one is just minus the other. Math 9C Summer 2011 (UCR) Pro-Notes June 25, 2011 1 / 13 The reason it’s called a telescoping series is because a lot of canceling occurs. Let’s look at the sequence of partial sums of a telescoping series of the first kind. s = ( a 1- a ) s 1 = ( a 1- a ) + ( a 2- a 1 ) s 2 = ( a 1- a ) + ( a 2- a 1 ) + ( a 3- a 2 ) . . . s n = ( a 1- a ) + ( a 2- a 1 ) + ( a 3- a 2 ) + ··· ( a n +1- a n ) After all the canceling, s n = a n +1- a . In other words ∞ X k =0 ( a k +1- a k ) = lim n →∞ s n = lim n →∞ a n +1- a . So a telescoping series ∞ X k =0 ( a k +1- a k ) converges if and only if { a k } converges . Math 9C Summer 2011 (UCR) Pro-Notes June 25, 2011 2 / 13 Example 1 Does the series ∞ X k =1 1 k + 1- 1 k converge? If so what does it sum to? You can immediately see this fits into the pattern of a telescoping series. Here a k = 1 / k . If you didn’t recognize this, you could try and write out a few terms of the sequence of partial sums and see if you noticed anything (you would have noticed all the canceling like before). Since { 1 / k } converges to 0, we know the series converges and ∞ X k =1 1 k + 1- 1 k = lim n →∞ 1 n + 1- 1 1 =- 1 Math 9C Summer 2011 (UCR) Pro-Notes June 25, 2011 3 / 13 Example 2...

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