1 求a1 a2
a[0]=1/2,a[n]=a[n-1]+a[n-1]^2/n^2
a[1]=a[0]+a[0]^2/1^2=1/2+1/4=3/4
a[2]=3/4+(3/4)^2/2^2=57/64
2 求證 1/a(n-1)-1/an<1/n^2
a[n]=a[n-1]+a[n-1]^2/n^2>a[n-1]
n^2(a[n]-a[n-1])=a[n-1]^2
所以1/a[n-1]-1/a[n]
=(a[n]-a[n-1])/a[n]a[n-1]
=a[n-1]/n^2[an]<1/n^2
即：1/a[n-1]-1/a[n]<1/n^2
得證.
3 求證 (n+1)/(n+2)數(shù)學(xué)歸納法：
n=1時,
a[1]代入得：(n+1)/(n+2)設(shè)n=k時,(k+1)/(k+2)則：n=k+1時,
a[k+1]=a[k]+a[k]^2/(k+1)^2
>(k+1)/(k+2)+1/(k+2)^2
=(k^2+3k+3)/(k+2)^2
(k^2+3k+3)/(k+2)^2>(k+2)/(k+3)
即：(k+2)^3<(k^2+3k+3) (k+3)
K^3+6k^2+12k+8所以(k+2)/(k+3)即對n=N,(n+1)/(n+2)故得證!