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  • $\begingroup$ Hi Jim. Thanks for your answer. I now used your NonlinearModelFit however I got a different a and b: {a -> 1.70003*10^-6, b -> 1.20435} $\endgroup$ Commented Feb 27, 2017 at 7:03
  • $\begingroup$ And yes, you are right it is not the best fit. I will try to make it better, but what I am looking for is just an asymptotic behaviour of the Fk, thus from the fit I already kind of see what to expect. $\endgroup$ Commented Feb 27, 2017 at 7:13
  • $\begingroup$ Using loga as the parameter rather than a in Log[a] gives a number much closer to yours and (maybe more importantly) gives a smaller error variance. However, if your objective is to estimate asymptotic behaviour and you're not so interested in small values of k, then that probably doesn't help much. $\endgroup$ Commented Feb 27, 2017 at 17:06
  • $\begingroup$ Yeah I realized that even for getting just an asymptotic behaviour my fit was far too bad, thus I changed the model to ak^(bk^c). Now the fit is much better and the residuals pretty small. Thanks again for your answer. $\endgroup$ Commented Feb 28, 2017 at 7:04