Continued fractions as inverted recurrence relations

[Don Reble]

Joseph Biberstine wrote:
> 4/Pi = 1 + 1/(3 + 4/(5 + 9/(7 + 16/(9 + 25/(11 + ...))))) that is, this 
> is a non-simple cont.frac. where the numerators are the squares and the 
> partial quotients are the odds.
> 
> Consider a[n] = 2*n - 1 + n^2/a[n+1] with a[1] = 4/Pi.

> What is the limit of a[n]/n as n->Infinity? 
> 
> If the limit does exist, it appear to be approximately 0.414 (roughly 
> a[2000]/2000), though I have no means to calculate it further.

    It's tricky to do this one correctly.

    Each a[n] can be represented as (a + b PI) / (c + d PI), for
    some integers a,b,c,d. These integers enlarge steadily as n
    increases; but the numerator and denominator approach zero. With
    finite precision, it doesn't take long for subtractive cancellation
    to render the calculation meaningless.

    I find that, when using D digits of precision, only about 1.3D terms
    can be computed. So I suspect that if Joseph uses only 1500 digits,
    his a[2000] is nonsense.

    I also find that after the accuracy goes (or if one uses a wrong
    value for Pi), a[n]/n appears to converge to -0.41421...
    Alas, that doesn't explain Joseph's +0.414.

    ---

    It may help to observe that a[n]/n decreases, while a[n]/(n+1)
    increases. Perhaps a[n]/(n+1/2) will be revealing.
    Using 2000 digits of precision, I find that that quotient
    reaches 2.41421356 before the accuracy goes. Maybe 1+sqrt(2) is the
    true limit?

> We suspect the value of a[1] is largely inconsequential, ...
> If this limit exists, how if at all does it relate to Pi? 

    I speculate, that 4/Pi is the only a[1] value, such
    that a[n]/n converges to 1+sqrt(2) instead of 1-sqrt(2).

-- 
Don Reble  djr at nk.ca


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