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**Unformatted text preview: **Infinite Series This is where the fun begins. If we have an infinite sequence { a k } k =0 we can get a brand new sequence { s n } n =0 by defining s = a s 1 = a + a 1 s 2 = a + a 1 + a 2 s 3 = a + a 1 + a 2 + a 3 . . . s n = a + a 1 + a 2 + + a n . . . Math 9C Summer 2011 (UCR) Pro-Notes June 23, 2011 1 / 16 The question we are really after is: What can we say about the new sequence { s k } . Does it converge or diverge? By saying it converges, it is the same thing as saying if we add more and more of the a k s together, were getting closer and closer to some number. Its strange to think that adding more and more things together would add up to anything at all. Math 9C Summer 2011 (UCR) Pro-Notes June 23, 2011 2 / 16 For example, Take the sequence { a k } = { (1 / 2) k } . Then the new sequence { s n } is s = 1 2 s 1 = 1 2 + 1 2 1 s 2 = 1 2 + 1 2 1 + 1 2 2 s 3 = 1 2 + 1 2 1 + 1 2 2 + 1 2 3 . . . s n = 1 2 + 1 2 1 + 1 2 2 + + 1 2 n . . . Math 9C Summer 2011 (UCR) Pro-Notes June 23, 2011 3 / 16 We can write down the first few terms of this sequence, i.e., { s , s 1 , s 2 , s 3 , .. } = { 1 , 3 / 2 , 7 / 4 , 15 / 8 , ... } Surprisingly, it looks like this sequence is converging. Remember that the n-th term is the sum of n things, so asking what happens as n goes to infinity is the same as asking what happens when we add more an more positive numbers together. Sounds like it should go to infinity, but it doesnt. Notice s n +1- s n = (1 / 2) n +1 > 0, so { s n } is increasing. Also s n < 2 (We always just add half the remaining distance to 2). By the bounded convergence theorem, { s n } converges. Do we know what it converges to?...

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