已知函数f(x)=ln( [e^x-1]/x)且数列{an}满足a1=1,an+1=f(an)求证{an}单调递减且an>0
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![已知函数f(x)=ln( [e^x-1]/x)且数列{an}满足a1=1,an+1=f(an)求证{an}单调递减且an>0](/uploads/image/z/5316690-66-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3Dln%EF%BC%88+%5Be%5Ex-1%5D%2Fx%29%E4%B8%94%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E6%BB%A1%E8%B6%B3a1%3D1%2Can%2B1%3Df%28an%29%E6%B1%82%E8%AF%81%EF%BD%9Ban%EF%BD%9D%E5%8D%95%E8%B0%83%E9%80%92%E5%87%8F%E4%B8%94an%3E0)
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