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   SIGMAarctan(2/n^2)
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   Author  Topic: SIGMAarctan(2/n^2)  (Read 1623 times)
ThudnBlunder
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SIGMAarctan(2/n^2)  
« on: May 20th, 2008, 5:37am »
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Evaluate tan-1(2/n2)  
           n=1
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Barukh
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Re: SIGMAarctan(2/n^2)  
« Reply #1 on: May 20th, 2008, 10:44am »
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135o
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ThudnBlunder
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Re: SIGMAarctan(2/n^2)  
« Reply #2 on: May 20th, 2008, 10:52am »
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on May 20th, 2008, 10:44am, Barukh wrote:
135o

Was that computer-assisted?   Roll Eyes
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Re: SIGMAarctan(2/n^2)  
« Reply #3 on: May 20th, 2008, 11:19am »
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on May 20th, 2008, 10:52am, ThudanBlunder wrote:

Was that computer-assisted? Roll Eyes

 
No.
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ThudnBlunder
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Re: SIGMAarctan(2/n^2)  
« Reply #4 on: May 20th, 2008, 11:27am »
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on May 20th, 2008, 11:19am, Barukh wrote:

 
No.

Then I'm beginning to believe our literary tastes are similar.   Tongue
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Re: SIGMAarctan(2/n^2)  
« Reply #5 on: May 20th, 2008, 11:14pm »
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hidden:
Solution is based on the following identity:
 
tan-1(2/n2) = tan-1(n+1) - tan-1(n-1)
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Eigenray
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Re: SIGMAarctan(2/n^2)  
« Reply #6 on: May 21st, 2008, 2:45am »
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Or less cleverly, working out the first few partial sums suggests
 
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arctan{-(n-1)(n+2)/[n(n-3)]} + arctan{2/n2} = arctan{-n(n+3)/[(n+1)(n-2)]}
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william wu
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Re: SIGMAarctan(2/n^2)  
« Reply #7 on: May 21st, 2008, 3:59am »
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Digression: As a knee jerk reaction, I took the derivative of the summand, and tried summing that instead. Not that that would lead to anything relevant for this problem ... but I ended up with something that surprised me:
 
d/dx [ArcTan[2/x^2]] = -(4 x)/(4 + x^4)
-(4 n)/(4 + n^4)  = -3/2
 
OK, now someone explain why I shouldn't be surprised Roll Eyes
« Last Edit: May 21st, 2008, 4:00am by william wu » IP Logged


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