Asymptotic equivalent of a sequence

**Hugal** · July 2nd 2011, 06:15 AM

Hi everyone,

I've a little trouble with a small exercice.
So, given the equation $x^n+x=1$ , it can ben easily shown that $\displaystyle \exists ! x_n \in \mathbb{R}^+ / x_n^n + x_n = 1$ .

I also prooved that the sequence $(x_n)_{n > 0}$ is convergent and its limit is $l=1$ .

Now, the question is : Find an asymptotic equivalent of $x_n - l$ . And this is what I have trouble with.

I've tried some thing, for example, we know that $x_n - 1 = - x_n^n$ but this expression do not seem to bring something good.
So, I don't want the whole answer, but just the little trick that I've not seen and which can end this question.

Thanks for reading me, and sorry if my english is little bit bad,

Hugo.

**girdav** · July 3rd 2011, 02:24 AM

Hi,
we have $x_n = 1-x_n^n = (1-x_n)\sum_{k=0}^{n-1}x_n^k$ hence $\displaystyle 1-x_n = \dfrac{x_n}{\sum_{k=0}^{n-1}x_n^k} = \dfrac{x_n(1-x_n)}{1-x_n^n}$ . Since $\lim_{n\to\infty}x_n=1$ we have $1-x_n\sim\frac{1-x_n}{1-x_n^n}$ . Now, use the mean value theorem to show that an asymptotic equivalent of $1-x_n$ is $\frac 1n$ .

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