Question: x1>0, xn+1=xn+1xn, n∈N. Evaluate
lim
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!
Answer
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.
No comments:
Post a Comment