令an=n!/(2^n+1)
lim(n->∞) a(n+1)/an
=lim(n->∞) {(n+1)!/[2^(n+1)+1]}/[n!/(2^n+1)]
=lim(n->∞) n(2^n+1)/[2^(n+1)+1]
=lim(n->∞) (2^n+1+n*ln2*2^n)/[ln2*2^(n+1)]
=lim(n->∞) (ln2*2^n+ln2*2^n+n*ln2*ln2*2^n)/[ln2*ln2*2^(n+1)]
=lim(n->∞) (2+n*ln2)/(2*ln2)
=∞
所以根据比值判别法，原级数发散
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