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x(1)>0

\( x(n+1)=\frac{x(n)}{1+n{x(n)}^2} \)

1/ Study the convergence of x(n).

2/ Find an equivalent of x(n).

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Maple gives:
\( x(n) = {n}^{-1}+cn^{-2}+c^{2}n^{-3}+c^{3}n^{-4}+O\left( {n}^{-5}\right) \)

o.m. wrote:x(1)>0

\( x(n+1)=\frac{x(n)}{1+n{x(n)}^2} \)

1/ Study the convergence of x(n).

2/ Find an equivalent of x(n).

For the first part:

1/ It is quite easy to see that from induction we get \( x_{n}>0, \forall n\geq 1 \). On the other hand, we have

\( \frac{x_{n+1}}{x_{n}}=\frac{1}{1+nx_{n}^{2}}<1, \)

so our sequence is decreasing, and by Weierstrass it is convergent.
But, from AM-GM inequality we get

\( x_{n+1}=\frac{x_{n}}{1+nx_{n}^{2}}\leq\frac{x_{n}}{2x_{n}\sqrt{n}}=\frac{1}{2\sqrt{n}} \).
Since \( x_{n}>0 \) by the sandwitch theorem it follows that \( \lim_{n\to\infty}x_{n}=0 \).