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Along the thread given here, Find the limit of a recursive sequence, I am very curious about the limit of the recursive sequence defined by \begin{gather*} u_0>0, \quad u_1>0,\quad u_{n+2}=\sqrt{u_{n+1}}+\sqrt{u_n},\quad \forall n\in\mathbb{N}. \end{gather*}

After some trial and error, I guess that there may be a generalization. I put my conjecture below. Let $k\in\mathbb{N}\backslash\{0,1\}.$ Assume \begin{align*} u_0>0,\quad u_1>0, \quad u_{n+2}=\sqrt[k]{u_{n+1}}+\sqrt[k]{u_{n}},\quad \forall n\in\mathbb{N}. \end{align*} Then we may have \begin{gather*} \lim_{n\to\infty} u_n=\sqrt[k-1]{2^k}=2^{\frac{k}{k-1}}. \end{gather*}

Is my conjecture true or false? Can you prove or disprove it?

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  • $\begingroup$ @BrainMScott, Sorry, I have trouble in editing, for there is always pop-up, saying "Mathematics Stack Exchange requires external JavaScript from another domain, which is blocked or failed to load." Therefore, I have to type very quickly, and so made some errors. $\endgroup$
    – azc
    Nov 4, 2014 at 5:42

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If the sequence converges, the limit must satisfy $u = \sqrt[k]{u} + \sqrt[k]{u}$, giving indeed $u = 2 ^ {\frac{k}{k-1}}$. Just prove that it does converge.

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