用比值审敛法或根值审敛法判别下列级数的收敛性：
(1)∑n=1∞3n/(n•2n)
(2)∑n=1∞(n•3n)
(3)∑n=1∞[n/(2n+1)]n)
(4)∑n=1∞[n/(3n-1)]2n
(5)∑n=1∞1/[ln/(3+1)]n
(6)∑n=1∞3n•n/nn
【正确答案】：(1)∵limn→∞(un+1)/un=limn→∞3n+1/(n+1)2n+1 •[(n•2)n/3n]=3/2﹥1 ∴∑n=1∞3n/(n•2n)发散 (2)∵limn→∞(un+1)/un=limn→∞(n+1)/3n+1•(3n/n)=1/3﹤1 ∴∑n=1∞n/3n收敛 (3)∵limn→∞n√un=limn→∞n/(2n+1)=1/2﹤1 ∴∑n=1∞[n/(2n+1)]n收敛 (4)∵limn→∞n√un=limn→∞[n/(3n+1)]2=1/9﹤1 ∴∑n=1∞[n/(3n+1)]2n收敛 (5)∵limn→∞n√un=limn→∞1/ln(n+1)=0 ∴∑n=1∞1/[ln(n+1)]n收敛 (6)∵limn→∞[3n+1•(n+1)!/(n+1)n+1]•[nn/3nn!] =limn→∞3(n+1)/(n+1)•(1+1/n)n=3e﹥1 ∴∑n=1∞(3n•n)/nn发散．
