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Unformatted text preview: Power Series John E. Gilbert, Heather Van Ligten, and Benni Goetz Whats the pay-off for introducing all these various tests and applying them to complicated series? Well, think of the remarkable series n = 0 x n n ! = 1 + x + x 2 2! + x 3 3! + x 4 4! + . . . + x n n ! + . . . = e x we mentioned right at the beginning. Of course, we still have no idea why the sum of the series is e x ; thats coming up soon! But at least now we see that the series converges absolutely for all x because by the Ratio test lim n a n +1 a n = lim n x n +1 x n n ! ( n + 1)! = lim n x n = 0 for each x . The series thus has a finite sum for each x , even if we havent actually determined the sum of the series or know why absolute convergence helps. Now lets look at Geometric series and Harmonic series in this context. Thinking of the common ratio as a variable x , not as a number r , we get f ( x ) = n = 0 x n = 1 + x + x 2 + x 3 + x 4 + . . . + x n + . . . = 1 1- x (take a = 1 for simplicity). The series converges absolutely on (- 1 , 1) and on this interval its sum is 1 1- x , so the series provides a series representation of a simple function from high school days. But this is calculus, so why not integrate both the Geometric series and its sum: x n = 0 t n dt = n = 0 x t n dt = n = 0 x n +1 n + 1 = n = 0 x n n x 1 1- t dt =- ln(1- x ) = ln 1 1- x , assuming that the integral of the infinite sum is the sum of the infinite number of integrals ; thats coming up too! Thus ln 1 1- x = x + x 2 2 + x 3 3 + x 4 4 + . . . + x n n + . . . , which on letting x 1- becomes ln( ) = 1 + 1 2 + 1 3 + 1 4 + . . . + 1 n + . . . = ....
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