Re: Pure probability question
Alternate solution in white:
<font color="white">Let f(x) be the expected number of terms before the sum is at least x. f(x) = 1 + Integral with respect to z of f(z) from z=x-1 to z=x. f'(x)=f(x)-f(x-1). f(x)=0 for -1<x<0, so for 0<x<1, f'(x)=f(x). The limit of f(x) as x decreases to 0 is 1, so by solving this differential equation with initial condition f(0)=1, f(x)=e^x on 0<x<=1. f(1)=e. </font>
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