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### Lecture44-rev

Course: MATH 10250, Fall 2008
School: Notre Dame
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Word Count: 252

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A26 MAT Lecture 44 1 Then Root Test for Convergence (The Root Test) Suppose that lim |ak |1/k = r exists. k k=1 k=1 proposition 1.1. r&lt;1 r&gt;1 r=1 ak converges absolutely ak diverges nothing at all (test is inconclusive in this case) proof. We the proof of the case |r| &lt; 1 only. Since |r| &lt; 1 we may choose s (r, 1). Note that 0 &lt; r &lt; 1 and then 0 &lt; |ak...

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A26 MAT Lecture 44 1 Then Root Test for Convergence (The Root Test) Suppose that lim |ak |1/k = r exists. k k=1 k=1 proposition 1.1. r<1 r>1 r=1 ak converges absolutely ak diverges nothing at all (test is inconclusive in this case) proof. We the proof of the case |r| < 1 only. Since |r| < 1 we may choose s (r, 1). Note that 0 < r < 1 and then 0 < |ak |1/k < s for k N (for some N ). Thus |ak | < sk for k N and so |ak | < k=N k=N k=N sk = sk 1-s So by the comparison test, k=1 ak . ak converges absolutely, and hence so does example 1.2. Does the root test apply to the series k=1 xk k! 1/k solution. We have |ak | The limit r = lim 1/k xk = k! = |x| (k!)1/k |x| can indeed be found as follows (Note: Our (k!)1/k k method here is by brute force; using the the ratio test is easier.) Assuming 1 k = 2 is even we have (k!)1/k = (1 2 3 ( + 1) k)1/k k = 2 factors (( + 1) k)1/k factors /k = ( 1 k) 2 2 On the other hand if k = 2 + 1 is odd, we have (k!)1/k = (1 2 3 ( + 1) k)1/k k = 2 + 1 factors (( + 1) k)1/k + 1 factors +1 ( + 1) k ( 1 k) 2 (since + 1 = 2 1 1 1 k+1 2 > 1k 2 1 +1 > ) k 2 1 In either case, (k!)1/k ( 2 k) 2 as k . Thus r = 0 and we have convergence. example 1.3. 1 . 2k+(-1)k k=1 See whether or not the the root test applies to the series solution. We have |ak | 1/k = 1 2k+(-1)k 1/k = 1 2(k+(-1)k )/k = 1 2 1 1+ k (-1)k 1 as k 2 Since lim |ak |1/k = k 1 < 1 the series converges. 2 2
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