```

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riddles >> putnam exam (pure math) >> SIGMAarctan(2/n^2)
(Message started by: ThudanBlunder on May 20th, 2008, 5:37am)

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Title: SIGMAarctan(2/n^2)
Post by ThudanBlunder on May 20th, 2008, 5:37am
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif
n=1

Title: Re: SIGMAarctan(2/n^2)
Post by Barukh on May 20th, 2008, 10:44am
[hide]135o[/hide]

Title: Re: SIGMAarctan(2/n^2)
Post by ThudanBlunder on May 20th, 2008, 10:52am

on 05/20/08 at 10:44:52, Barukh wrote:
 [hide]135o[/hide]

Was that computer-assisted?   ::)

Title: Re: SIGMAarctan(2/n^2)
Post by Barukh on May 20th, 2008, 11:19am

on 05/20/08 at 10:52:59, ThudanBlunder wrote:
 Was that computer-assisted?   ::)

No.

Title: Re: SIGMAarctan(2/n^2)
Post by ThudanBlunder on May 20th, 2008, 11:27am

on 05/20/08 at 11:19:12, Barukh wrote:
 No.

Then I'm beginning to believe our literary tastes are similar.   :P

Title: Re: SIGMAarctan(2/n^2)
Post by Barukh on May 20th, 2008, 11:14pm
[hideb]Solution is based on the following identity:

tan-1(2/n2) = tan-1(n+1) - tan-1(n-1)[/hideb]

Title: Re: SIGMAarctan(2/n^2)
Post by Eigenray on May 21st, 2008, 2:45am
Or less cleverly, working out the first few partial sums suggests

[hideb]arctan{-(n-1)(n+2)/[n(n-3)]} + arctan{2/n2} = arctan{-n(n+3)/[(n+1)(n-2)]}[/hideb]

Title: Re: SIGMAarctan(2/n^2)
Post by william wu on May 21st, 2008, 3:59am
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)
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/csigma.gif-(4 n)/(4 + n^4)  = -3/2

OK, now someone explain why I shouldn't be surprised ::)