an+1=a1+∑ni=1(ai/i)2, 可以視為下列定積分,
∫nt=1(x(t)/t)2dt=x(t)= a1t/[a1+(1-a1)t],
根據洛必達法則求極限,得到,當n和t趨於無窮大時:
lim an+1=lim(x(t)) = a1/(1-a1) ,其中0<=a1<1。
an+1=a1+∑ni=1(ai/i)2, 可以視為下列定積分,
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