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Topic: Convergent Series (Read 718 times) 

ThudnBlunder
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Convergent Series
« on: Apr 1^{st}, 2007, 5:20pm » 
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For what range of a > 0 does the series below converge? a^{(1 + 1/2 + 1/3 + ....... + 1/n)} n=1


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iyerkri
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Re: Convergent Series
« Reply #1 on: Apr 1^{st}, 2007, 8:42pm » 
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My answer : a < 1/e .


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Icarus
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Re: Convergent Series
« Reply #2 on: Apr 2^{nd}, 2007, 5:05pm » 
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Providing details: Let L(n) = 1 + 1/2 + ... +1/n. Obviously, the series diverges for a 1, so assume a < 1. ln(n) < L(n) < ln(n) + , so a^{} a^{ln(n)} a^{L(n)} a^{ln(n)}. Hence the series converges or diverges with a^{ln(n)}. But a^{ln(n)} = n^{ln(a)}, and n^{ln(a)} is well known to converge if and only if ln(a) < 1. Hence a^{L(n)} converges if and only if 0 < a < 1/e.

« Last Edit: Apr 2^{nd}, 2007, 5:15pm by Icarus » 
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"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? "  Anonymous



