设an=n[2^(1/n)-1]
lim(n->∞) an
=lim(n->∞) [2^(1/n)-1]/(1/n)
=lim(x->0) (2^x-1)/x
=lim(x->0) (2^xln2)/1
=ln2
≠0
所以∑an是发散的.