19/03/22 15:02:21.28 WSdp8+VY.net
>>696 補足
γ＝lim(n→∞)1+1/2+1/3+…+1/n-log(n+α)≡lim(n→∞)γn(α)
In particular, γn(1/2) exhibits much more rapid convergence than the conventional expansion γn(0).
で、γn(1/2) つまりα＝1/2がベストかね？ はて？（＾＾
19/03/22 15:02:21.28 WSdp8+VY.net
>>696 補足
γ＝lim(n→∞)1+1/2+1/3+…+1/n-log(n+α)≡lim(n→∞)γn(α)
In particular, γn(1/2) exhibits much more rapid convergence than the conventional expansion γn(0).
で、γn(1/2) つまりα＝1/2がベストかね？ はて？（＾＾