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I got the following result : $x_{n+1} = x_{n}+\frac{1}{x_{n}}$ and so forth.

As $x_{1}>0$, then I know by plugging that $x_{k}>1$ for every natural $k$. Hence, the terms in the denominator tend to infinity as $n$ tends to infinity, which left $x_{n+1} = x_{n}$, so it is a constant sequence and hence converges. $x_{n+1} = x_{n}$ Is it really right?

Thanks in advance.

Gary
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Victor Hugo
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    If it converges to some value $L > 0$, then it should satisfy $L = L + 1/L$, which is impossible. – Seewoo Lee Apr 28 '22 at 05:40
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    It is increasing and cannot tend to finite limit so it must tend to $\infty$. – Kavi Rama Murthy Apr 28 '22 at 05:41
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    Note that $x_k$ is monotonically increasing. Then it either converges to infinity or to a fixed positive number. But the second case is not possible by Lee's comment. – stephenkk Apr 28 '22 at 05:41
  • If the terms in the denominator (i.e., $x_n$) tend to infinity (as you claim) then it literally means that $x_n$ tends to infinity. Thus, it cannot be a constant sequence. You cannot just replace $x_n$ by its limit in some places and leave it as $x_{n+1}$ and $x_n$ in others. Also, how exatly did you show that $x_n$ tends to infitiy? Monotonicity does not guarentee that. – Gary Apr 28 '22 at 06:01

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You first show $x_n > 0$ for all $n$ by induction. Next square both sides of the equation: $x_n^2 > x_{n-1}^2 + 2$. Thus $x_n^2 > 2(n-1)\implies x_n > \sqrt{2(n-1)}\implies x_n \to \infty$.

Wang YeFei
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