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Topic: Evaluate Sum (Read 1699 times) |
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ThudnBlunder
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Evaluate Sum
« on: Jul 22nd, 2010, 5:55pm » |
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Evaluate for k = 1 to   1/k(n + 1)
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Immanuel_Bonfils
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Re: Evaluate Sum
« Reply #1 on: Jul 23rd, 2010, 5:41am » |
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Is n+1 a constant factor in the denominator?
In that case it 'll divide an harmonic series that diverges to + \infinity (I couldn't get to the symbos "facilities")
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0.999...
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Re: Evaluate Sum
« Reply #2 on: Jul 23rd, 2010, 5:46am » |
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If instead we are summing over 1/(k^2+k) then the sum will be 1 after making it into a telescoping series.
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ThudnBlunder
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Re: Evaluate Sum
« Reply #3 on: Jul 23rd, 2010, 3:17pm » |
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Sorry, major typo! That should have been:
Evaluate for k = 1 to 1/[k(n + k)]
Stilll not that difficult.
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0.999...
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Re: Evaluate Sum
« Reply #4 on: Jul 24th, 2010, 4:23pm » |
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It is easily seen that by the comparison test the sum is absolutely convergent for n >= 0.
For n = 0, it evaluates as a special case (AFAIK) to  2/6
For n > 0, we shall split each term into 1/n(1/k-1/(n+k)). And since the series is absolutely convergent, we may rearrange the terms like so:
1/n[H(n) + (1/(n+1)-1/(n+1))+(1/(n+2)-1/(n+2))+...]
= 1/n*H(n) where H(n) is the harmonic sum from 1 to n.
In the scope of the problem, I do not see a means of further simplifying the result.
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| « Last Edit: Jul 24th, 2010, 7:48pm by 0.999... » |
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