答：
limn->∞ u(n+1)/u(n)
=limn->∞ [(n+1)tan(π/2^(n+2))]/[ntan(π/2^(n+1))]
又当t->0时,tant~t
=limn->∞ [(n+1)(π/2^(n+2))]/[n(π/2^(n+1))]
=limn->∞ (n+1)/(n*2)
=1/2
判定级数ntan (π\2∧n+1)的敛散性
答：
limn->∞ u(n+1)/u(n)
=limn->∞ [(n+1)tan(π/2^(n+2))]/[ntan(π/2^(n+1))]
又当t->0时,tant~t
=limn->∞ [(n+1)(π/2^(n+2))]/[n(π/2^(n+1))]
=limn->∞ (n+1)/(n*2)
=1/2