将函数f(x)=lnx展开成x-1的幂级数
可以简单推导一下：
1/(1-x) = 1+x+x^2+…+x^n+…
integral from 0 to x,
ln(1-x) = x+x^2/2+…+x^n/n+…
lnx = ln(1-(1-x)) = (1-x)+(1-x)^2/2 + … + (1-x)^n/n + …
Answer: lnx = -(x-1)+(x-1)^2/2 + …+ (-1)^n(x-1)^n/n+…, n from 1 to infinity
可以简单推导一下：
1/(1-x) = 1+x+x^2+…+x^n+…
integral from 0 to x,
ln(1-x) = x+x^2/2+…+x^n/n+…
lnx = ln(1-(1-x)) = (1-x)+(1-x)^2/2 + … + (1-x)^n/n + …
Answer: lnx = -(x-1)+(x-1)^2/2 + …+ (-1)^n(x-1)^n/n+…, n from 1 to infinity