Modification on a sequence from VJIMC

The first two limits of the following problem were proposed at VJIMC, 2005, Category I.

The exact value of the last limit was proposed by a user at https://artofproblemsolving.com , where you can also see his solution along other lines.

 Let (x_n)_{n\ge 2} be a sequence of real numbers, such that x_2>0 and for every n\ge 2 holds 

x_{n+1}=-1+\sqrt[n]{1+nx_n}.


Prove consecutively that  1)\ \lim_{n\to\infty}x_n=0,\  \ \ 2)\ \lim_{n\to\infty}nx_n=0,\  \ \ 3)\ \lim_{n\to\infty}n^2x_n=4.


\textbf{Proof.}  \textbf{1)} Clearly all the elements of the sequence are positive. The inequality -1+\sqrt[n]{1+nx_n}<x_n is equivalent to (1+nx_n)<(1+x_n)^n, which is seen to be true after expanding, since all the summands on the right are positive. This shows that the sequence is strictly decreasing. Hence

0<x_{n+1}=-1+\sqrt[n]{1+nx_n}\le -1+\sqrt[n]{1+nx_2},

and since the right hand side clearly tends to 0 we obtain \displaystyle\lim_{n\to\infty}x_n=0.

\textbf{2)} Now 1/x_n\to +\infty  increasingly and we can use Stolz theorem as follows:

\lim_{n\to\infty}nx_n =\lim_{n\to\infty}\frac{n}{\frac{1}{x_n}}=\frac{(n+1)-n}{\frac{1}{x_{n+1}}-\frac{1}{x_n}}=\lim_{n\to\infty}\frac{x_{n+1}x_n}{x_n-x_{n+1}}

Now consider the defining equation. It can be rewritten as (1+x_{n+1})^n=1+n x_n hence x_n=x_{n+1}+S/n where \displaystyle S=\sum_{k=2}^n{n\choose k}x_{n+1}^k.

Thus 

\lim_{n\to\infty}\frac{x_{n+1}x_n}{x_n-x_{n+1}}=\lim_{n\to\infty}\frac{x_{n+1}\left(x_{n+1}+\frac{S}{n}\right)}{\frac{S}{n}}=\lim_{n\to\infty}\frac{nx_{n+1}^2}{S},

where we have used that x_{n+1}\to 0. Now this can be rewritten as follows

\lim_{n\to\infty}\frac{nx_{n+1}^2}{S}=\lim_{n\to\infty}\frac{n}{{n\choose 2}+S'},

 where \displaystyle S'=\sum_{k=3}^n{n\choose k}x_{n+1}^{k-2}>0. This shows that the last limit is 0.

\textbf{3)}  Apply the Stolz theorem in the very same way as above -


\lim_{n\to\infty}n^2x_n =\lim_{n\to\infty}\frac{n^2}{\frac{1}{x_n}}=\frac{(n+1)^2-n^2}{\frac{1}{x_{n+1}}-\frac{1}{x_n}}=\lim_{n\to\infty}(2n+1)\frac{x_{n+1}x_n}{x_n-x_{n+1}}=2\lim_{n\to\infty}\frac{n^2x_{n+1}^2}{S}.

For convenience we work with the reciprocal limit 

\lim_{n\to\infty}\frac{S}{n^2x_{n+1}^2}=\lim_{n\to\infty}\frac{\sum_{k=2}^n{n\choose k}x_{n+1}^k}{n^2x_{n+1}^2}=\frac{1}{2}+\lim_{n\to\infty}\sum_{k=3}^n{n\choose k}\frac{1}{n^2}x_{n+1}^{k-2}.

 We would be done if we prove, that the last limit is 0.  Denote A_n=\displaystyle \sum_{k=3}^n{n\choose k}\frac{1}{n^2}x_{n+1}^{k-2}. Clearly A_n>0. Fix \varepsilon>0. According to 2), for large enough n holds \displaystyle x_{n+1}<\frac{\varepsilon}{n}.

Thus (after completing to the binomial formula) A_n<\sum_{k=3}^n{n\choose k}\frac{\varepsilon^{k-2}}{n^k}=\frac{-n \varepsilon ^2+2 n \left(\frac{n+\varepsilon }{n}\right)^n-2 n \varepsilon -2 n+\varepsilon ^2}{2 n \varepsilon ^2},

hence \limsup_{n\to\infty}A_n\le \lim_{n\to\infty} \frac{-n \varepsilon ^2+2 n \left(\frac{n+\varepsilon }{n}\right)^n-2 n \varepsilon -2 n+\varepsilon ^2}{2 n \varepsilon ^2}=\frac{-\varepsilon ^2-2 \varepsilon +2 e^{\varepsilon }-2}{2 \varepsilon ^2}.

This is true for arbitrary \varepsilon>0. Letting \varepsilon\to 0 in the last bound we obtain (using Taylor expansion of the exponent near 0) that  

\lim_{\varepsilon\to 0}\frac{-\varepsilon ^2-2 \varepsilon +2 e^{\varepsilon }-2}{2 \varepsilon ^2}=0,

whence  \displaystyle\limsup_{n\to\infty}A_n=0, which finishes the proof.

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