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For every $n \in N^*$, $f_n$ is the function defined on $\left]0;+\infty\right[$ as $f_n(x)= x^n+9x^2-4$. We've proven that for all $n \in N^*$, there exists only one single $\alpha_n\in \left]0;+\infty\right[$ such as $\!f_n(\alpha_n)= 0$ . How can I prove that $(\forall n\in N^*) \frac{2}{3}-\frac{1}{6}(\frac{2}{3})^n<\alpha_n<\frac{2}{3}$ ? Thanks a lot for your help !

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    That is pretty difficult to read. – 2017-02-03
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    I'm really sorry, I'm new here. What is to code for adding spaces please ? – 2017-02-03
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    MathJax seems not to be intended to accommodate the French notation for open intervals. Observe: $$\alpha_n\in]0;+\infty[$$ $$\text{versus}$$ $$ \alpha_n\in\left]0;+\infty\right\[$$ The latter has proper spacing and requires \left and \right$\,.$ The binary relation symbol $\in$ is treated as if nothing comes after it, and hence proper spacing isn't there, when the coding is as in the first example above. – 2017-02-03
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    Absolutely! Thanks a lot ! – 2017-02-03
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    One peculiarity of the English language is that "series" is both the singular and the plural. "This series is" and "These series are" are both correct. I edited accordingly. – 2017-02-03
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    I'm sorry for bothering you. – 2017-02-03

2 Answers 2

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$$ \begin{align} \alpha_{\small n}\gt0 &\Rightarrow \color{blue}{2}\lt(2+3\,\alpha_{\small n}) \\[2mm] f_{\small n}(\alpha_{\small n}) &= \alpha_{\small n}^{n}+9\,\alpha_{\small n}^2-4=0 \space\Rightarrow\space 4-9\,\alpha_{\small n}^2=(2+3\,\alpha_{\small n})(2-3\,\alpha_{\small n})=\alpha_{\small n}^{n}\gt0 \space\Rightarrow\space \color{red}{\alpha_{\small n}\lt\frac{2}{3}} \\[2mm] &\Rightarrow \color{blue}{2}\,(2-3\,\alpha_{\small n})\lt(2+3\,\alpha_{\small n})(2-3\,\alpha_{\small n})=\alpha_{\small n}^{n}\lt\left(\frac{2}{3}\right)^{n} \space\Rightarrow\space \color{red}{\alpha_{\small n}\gt\frac{2}{3}-\frac{1}{6}\,\left(\frac{2}{3}\right)^{n}} \\[2mm] \end{align} $$

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Hint: Show that $\beta_n := \frac{2}{3}-\frac{1}{6}(\frac{2}{3})^n$ and $\gamma := \frac{2}{3}$ satisfy $$ f_n(\beta_n) < 0 \text{ and } f_n(\gamma) > 0 \, . $$ Then $$ f_n(\beta_n) < f(\alpha_n) < f_n(\gamma) $$ and since $f_n$ is strictly increasing on $(0, \infty)$, this implies that $\beta_n < \alpha_n < \gamma$.