[seqfan] A very fast convergent alternating series for phi
Vladimir Shevelev
shevelev at bgu.ac.il
Sun Feb 24 16:36:56 CET 2013
Dear SeqFans,
I submitted a very fast convergent alternating series for phi in A001622:
phi=1+sum{k>=1}(-1)^(k-1) /(F(k)*F(k+1)),
where F(n) is the n-th Fibonacci number (A000045). Proof. By Catalan's identity,
F^2(n)-F(n-1)*F(n+1)=(-1)^(n-1).
Therefore, (-1)^(n-1)/(F(n)*F(n+1))=F(n)/F(n+1)-F(n-1)/F(n).
Thus sum{k=1,...,n}(-1)^(k-1) /(F(k)*F(k+1))=F(n)/F(n+1).
If n goes to infinity, then F(n)/F(n+1) tends to 1/phi=phi-1.
It is almost the best approxomation of irrational 1/phi by a rational numbers. Indeed, our formula yields 1/phi=F(n)/F(n+1)+R_n, where |R_n|<=1/(F(n+1)*F(n+2)). By Hurwitz's theorem there exist infinitely many rationals m/n with approximation (in particular, for phi) <1/(sqrt(5)*n^2) and the constant sqrt(5) is not improvable. If n:=F(n+1), then it is approximation of degree 1/(sqrt(5)*F^2(n+1))<1/(F(n+1)*F(n+2)), or the same sqrt(5)*F(n+1))>F(n+2) which essentially follows from sqrt(5)>phi.
Best regards,
Vladimir
Shevelev Vladimir
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