11.2.1-eg-1 - -1 2 1 , a 3 = 5 -1 2 2 , ..., which means...

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Example Consider the following series: 5 - 5 2 + 5 4 - 5 8 + · · · . (a) Give the value of the nth term a n that allows us to write the series as X n =1 a n . (b) What is the sum of the series? Solution: (a) In order to determine a n , find a pattern among the terms of the series: 5 - 5 2 + 5 4 - 5 8 + ··· = 5 · 1 1 - 5 · 1 2 + 5 · 1 4 - 5 · 1 8 + ··· = 5 · 1 2 0 - 5 · 1 2 1 + 5 · 1 2 2 - 5 · 1 2 3 + ··· = 5 · ( - 1) 0 2 0 + 5 · ( - 1) 1 2 1 + 5 · ( - 1) 2 2 2 + 5 · ( - 1) 3 2 3 + ··· = 5 · ± - 1 2 ² 0 + 5 · ± - 1 2 ² 1 + 5 · ± - 1 2 ² 2 + 5 · ± - 1 2 ² 3 + ··· To write this summation as X n =1 a n we thus need to identify a 1 = 5 · ± - 1 2 ² 0 , a 2 = 5 · ±
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Unformatted text preview: -1 2 1 , a 3 = 5 -1 2 2 , ..., which means that a n = 5 -1 2 n-1 , for n = 1 , 2 ,.... (b) Note that this series X n =1 5 -1 2 n-1 is in the form of a geometric series X n =1 ar n-1 , where a = 5 and r =-1 2 . Since | r | < 1, this series must converge. Further, its sum is given by a 1-r = 5 1-(-1 2 ) = 5 3 2 = 10 3 ....
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