a(n+1)=(n+1)ba(n)/[a(n)+n],若a(n+1)=0,则a(n)=0, ..., a(1)=0,与a(1)=b>0矛盾。因此，a(n)不为0。1/a(n+1) = [a(n)+n]/[(n+1)ba(n)] = [n/(n+1)b][1/a(n)] + 1/[(n+1)b],(n+1)/a(n+1)=[1/b][n/a(n)] + [1/b],若b=1,则 (n+1)/a(n+1) = n/a(n) + 1,{n/a(n)}是首项为1/a(1)=1/b=1,公差为1的等差数列。n/a(n) = 1 + (n-1) = n, a(n) = 1.若b>0,b不为1，则(n+1)/a(n+1) ...
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