求数列极限

${x}_{n}$=$\sqrt{a+\sqrt{a+\cdots\cdots+\sqrt{a}}}$(n个根式),$a>0$,$n=1,2,\cdots\cdots$极限存在,并求$\lim\limits_{n\to \infty}$${x}_{n}$.
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代数龙

赞同来自: 数分龙 mumutiancheng

In fact, the following paper have a genaralization proposition: A. Herschfeld, $On$ $infinite$ $radicals$. American Mathematical Monthly, 1935 419-429, that is
\begin{equation}\text{A sequence } u_n=\sqrt{a_1+\sqrt{a_2+\cdots+\sqrt{a_n}}} \text{is convergent}\displaystyle
\Leftrightarrow \varlimsup\limits_{n \rightarrow \infty} a_n^{2^{^{-n}}}<+\infty \end{equation}
It is obvious that $\varlimsup\limits_{n \rightarrow \infty} a^{2^{^{-n}}}=1<+\infty$, So we may assume that $\lim\limits_{x\to +\infty}x_n=x$, we can pass to limit $n \to +\infty$ in $x_n=\sqrt{a+x_{n-1}}$, giving $x=\sqrt{a+x}$ , thus $x=\dfrac{1+\sqrt{1+4a}}{2}$.
Now, I'll give you a elementary solution.
Since $a>0, x_1=\sqrt{a}<\sqrt{a+\sqrt{a}}=x_2$. By induction, if $x_{n}<x_{n+1}$, then $x_{n+1}=\sqrt{a+x_{n}}<\sqrt{a+x_{n+1}}=x_{n+2}$. Thus, $x_{n}$ is an increasing sequence.
To show the sequence $x_{n}$ is bounded, firstly, we consider the case where $a\geqslant 2$, in this case, $0<x_1=\sqrt{a}\leqslant a$, now if $0<x_n\leqslant a$ for any positive integer $n$, it follows that $0<x_{n+1}=\sqrt{a+x_{n}}\leqslant \sqrt{2a}\leqslant a$. Thus also by induction we have $0<x_{n}\leqslant 2$ for all $n$.
lastly, considering the case $0<a<2$, you can show that $0<x_n <2$ by analogous methods ( By induction).
Therefore, we have shown $x_n$ is monotone and bounded sequence.
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