Friday, 18 October 2013

real analysis - Limit of (xn) with $0




Let 0<x1<1 and xn+1=xnxn+1n for n.



Prove that the limit exists and find the limit in terms of x_1.




I have proved the existence but cannot manage the other part.




Thanks for any help.


Answer



You can clearly write each term x_{n} as a polynomial in x=x_1, and it should be apparent that all such polynomials are x + O(x^2). Then the difference x_{n+1}-x_{n}=x_{n}^{n+1} is O(x^{n+1}), and so the coefficient of x^{k} is the same for all x_{n} with n\ge k. This allows us to determine the power series expansion of x_\infty=\lim_{n\rightarrow\infty}x_{n}: it is
x_\infty(x) = x-x^2-x^3+2x^4+3x^6-20x^7+30x^8-11x^9-31x^{10}+228x^{11}+\dots
The OEIS doesn't have anything matching this particular sequence.


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