1/(1+x) = ∑(-x)^n, n=0,1,...
u(x)=1/(1+x/2) = ∑(-x)^n/2^n, n=0,1,...
u' = -∑n(-x)^(n-1)/2^n, n=0,1,...
u'' = ∑n(n-1)(-x)^(n-2)/2^n, n=0,1,...
=∑(n^2-n) (-x)^(n-2)/2^n, n=0,1,..
很显然，u''(x) +u(x) 当x=1等于所求极限
而u''(x)和u(x)在x=1处的函数值很容易求
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