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Question: $x_1>0$, $x_{n+1}=x_n+\dfrac1{x_n}$, $n\in\Bbb N$. Evaluate $$\lim_{n\to\infty}\frac{x_n}{\sqrt n}.$$

What I know now is that $\dfrac1{x_n}\to\dfrac12$ when $n\ge2$, $\{x_n\}$ is monotonically increasing，$x_n\ge 2$ when $n\ge 2$.

I have tried to use the Stolz theorem, and I found I could not use Squeeze theorem.

Could you please give some instructions? Thank you!

anmo
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2 Answers2

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Hint. Your are not correct, because $x_n\to+\infty$. Moreover, we have that $$\frac{x_{n+1}-x_n}{\sqrt{n+1}-\sqrt{n}}=\frac{\sqrt{n+1}+\sqrt{n}}{x_n}.$$ So, by Stolz-Cesaro, if $\lim_{n\to\infty}\frac{x_n}{\sqrt n}=L$, then $$L=\lim_{n\to\infty}\frac{x_n}{\sqrt n}=\lim_{n\to\infty} \frac{x_{n+1}-x_n}{\sqrt{n+1}-\sqrt{n}}=\lim_{n\to\infty}\frac{\sqrt{n+1}+\sqrt{n}}{x_n}=\frac{2}{L}.$$

Robert Z
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  • I got it.Thanks,Mr.Robert.Z, I found your approach is similar to @Tianlalu fundamentally. You all used a certain value to subsitute the limit of that expression, and I can get the answer just by solving an equation! Thank you for you instruction again! – anmo Oct 24 '18 at 06:59
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We have $$x_{n+1}^2=\left(x_n+\frac1{x_n}\right)^2=x_n^2+\frac1{x_n^2}+2\implies x_{n+1}^2-x_n^2=\frac1{x_n^2}+2.$$

Obviously, $x_n$ is increasing and $x_n\to\infty$ as $n\to\infty$. Apply the Stolz theorem, \begin{align*} \left(\lim_{n\to\infty}\frac{x_n}{\sqrt n}\right)^2&=\lim_{n\to\infty}\frac{x_n^2}{n}\\ (\text{Stolz})&=\lim_{n\to\infty}\frac{x_n^2-x_{n-1}^2}{n-(n-1)}\\ &=\lim_{n\to\infty}\left(\frac1{x_{n-1}^2}+2\right)=0+2=2. \end{align*} $$\therefore \lim_{n\to\infty}\frac{x_n}{\sqrt n}=\sqrt 2.$$

Tianlalu
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  • Thank you! Thank you for telling me a important way to solve such problem is to square the expression. – anmo Oct 24 '18 at 06:54