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 Sum ArcTan Limit (Posted on 2016-05-11)
F(k) denotes the kth Fibonacci Number.

Define:
G(n) = Σk=1 to n ArcTan((F(2k+1)-1)

Find:
Limit G(n)
n → ∞

 No Solution Yet Submitted by K Sengupta Rating: 4.0000 (1 votes)

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 Analytic Solution Comment 6 of 6 |
Dawn has broken at several levels.

Using the Fibonacci version with F(1) = F(2) = 1:

(F(2k + 1))2 = 1 + F(2k)F(2k+2)           (Catalan et al)

So  (F(2k + 1))-1 = F(2k + 1)/(F(2k + 1))2

= [F(2k + 2) – F(2k)]/[ 1 + F(2k)F(2k+2)]

Now using   [tan A - tan B]/[1 + tan A tan B] = tan(A - B):

arctan((F(2k + 1))-1) = arctan F(2k + 2) – arctan F(2k)

Summing from k=1 to k=n (and noting the cancellation of
successive terms on the RHS, with only two remnants):

G(n) = arctan F(2n + 2)  -  arctan F(2)

= arctan F(2n + 2)  – pi/4

In the limit as n -> infinity, F(2n + 2) -> infinity, and this
gives
G(n) ->  pi/2  -  pi/4  =  pi/4

 Posted by Harry on 2016-05-19 11:26:19

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