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Sum ArcTan Limit (Posted on 2016-05-11)

F(k) denotes the kth Fibonacci Number.

Define:
G(n) = Σ_{k=1 to n} ArcTan((F(2k+1)^-1)

Find:
Limit G(n)
n → ∞

See The Solution Submitted by K Sengupta

Rating: 4.5000 (2 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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