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        riddles >> easy >> Evaluate Sum
        
(Message started by: ThudanBlunder on Jul 22^nd, 2010, 5:55pm)

Title: Evaluate Sum
Post by ThudanBlunder on Jul 22^nd, 2010, 5:55pm

Evaluate for k = 1 to http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif1/k(n + 1)

Title: Re: Evaluate Sum
Post by Immanuel_Bonfils on Jul 23^rd, 2010, 5:41am

Is n+1 a constant factor in the denominator?
[hide]In that case it 'll divide an harmonic series that diverges to + \infinity (I couldn't get to the symbos "facilities")[/hide]

Title: Re: Evaluate Sum
Post by 0.999... on Jul 23^rd, 2010, 5:46am

If instead we are summing over 1/(k^2+k) then the sum will be [hide]1 after making it into a telescoping series[/hide].

Title: Re: Evaluate Sum
Post by ThudanBlunder on Jul 23^rd, 2010, 3:17pm

Sorry, major typo! That should have been:

Evaluate for k = 1 to http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif1/[k(n + k)]

Stilll not that difficult.

Title: Re: Evaluate Sum
Post by 0.999... on Jul 24^th, 2010, 4:23pm

[hide]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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif²/6
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.
[/hide]