
Re: Sequence limit
Posted:
Oct 26, 2013 5:17 AM


Am 26.10.2013 09:51, schrieb karl: > Am 26.10.2013 09:13, schrieb Roland Franzius: >> Am 05.10.2013 09:04, schrieb quasi: >>> Roland Franzius wrote: >>>> quasi wrote: >>>>> >>>>> How do you justify the claim >>>>> >>>>> abs(sin(eps_j)) < (1/2)*eps_j >>>>> >>>>> ?? >>>> >>>> We called it the epsilon trick. >>> >>> Definitely quite a trick since, for 0 < eps_j < Pi/2, the above >>> inequality is always false. >> >> >> The conjecture that liminf_{n > oo} sin n^(1/n) = 1 seems more or less plausible from computer experiments. >> >> Using Mathematica, I just checked the first 50 continued fraction approximations of pi which give the series of best >> rational bounds of pi within one unit of the denominator >> >> cf(pi,n) = a_n/b_n >> >> then >> >> log a_n is prop 1.15 n >> >> and >> >> >> log(sin a_n) is prop  1.15 n >> >> and for n<50 >> >> 0.5 < a_n * sin a_n < 1.5 >> >> For n greater than 55 my system with 8 GB gives up. >> >> Of course these experiments do not rule out, that very good approximations of pi occur infinitely often, but to me it >> seems more or less improbable that a subseries a_n_k of approximations giving >> >> liminf sin(a_n_k)^1/(a_n_k) < 1 >> >> may exist. >> >> >> See more on >> >> http://mathworld.wolfram.com/PiContinuedFraction.html >> >> for the series q_n ^(1/n) for the denominators q_n with a seemingly constant limit which according to Eric Weisstein >> has not yet proven. >> >> Here n is the index of the nth denominator in the truncated continued fraction series. After complete cancellation of >> common factors, the denominators and numerators grow exponentially with the index. >> >> > This is no answer to quasi's question, clever Dick! >
We don't give answers to open problems here, poor charly, have fun yourself.

Roland Franzius

