Question

求数学帝，怎么判断级数的敛散性？？？？

Answer 1

记a[n] = n!/n^n·2^n·sin(π/3^n).
则lim{n → ∞} a[n+1]/a[n]
= lim (n+1)!/n!·n^n/(n+1)^(n+1)·2^(n+1)/2^n·sin(π/3^(n+1))/sin(π/3^n)
= lim n^n/(n+1)^n·2·sin(π/3^(n+1))/(π/3^(n+1))·(π/3^n)/sin(π/3^n)·1/3
= 2/3·1/ lim (1+1/n)^n · lim sin(π/3^(n+1))/(π/3^(n+1)) / lim sin(π/3^n)/(π/3^n)
= 2/3·1/e·1/1
= 2/(3e) < 1,
根据比值判别法, 级数收敛(本身就是正项级数, 因此也是绝对收敛).

注: 其中用到n → ∞时π/3^(n+1) → 0,
而lim{x → 0} sin(x)/x = 1, 故lim{n → ∞} sin(π/3^(n+1))/(π/3^(n+1)) = 1.
同理故lim{n → ∞} sin(π/3^n)/(π/3^n) = 1.