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Let $x_1:=a>0$ and $x_{n+1}:=x_n+1/x_n$ for $n\in\mathbb{N}$. Determine whether $(x_n)$ converges or diverges.

My answer: $(x_n)$ is divergent.

Proof: Assume that $(x_n)$ converges to $x$. Then $\lim (x_{n+1})=\lim (x_n)$. That is, $x=x+1/x$. This equation has no solution. Hence, $(x_n)$ is not convergent. Therefore it is divergent.

My question. Is the proof above a legitimate one? Something just doesn't feel right to me. Or should I just show that the sequence is unbounded. Or am I altogether wrong?

Thank you.

nonremovable
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2 Answers2

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Your proof is fine.

Note that divergent simply means that it is not convergent, not that it goes to $\infty$.

lhf
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I can prove the stronger result: the sequence is unbounded and tends to $\infty$. It is given that $x_1=a>0$, so the first term is positive. By looking at the recurrence relation, it is clear that the entire sequence is positive ( can be proven easily by induction). Now, assume the sequence is bounded, so $x_n<L$ for all $n$ and some positive $L$. Hence, since $\frac{1}{n}$ is strictly decreasing for positive $n$, we have that $\frac{1}{x_n}>\frac{1}{L}$. Hence $$ x_{n+1} = x_{n} + \frac{1}{x_n} > x_{n} + \frac{1}{L} $$ Hence, repeatedly applying this inequality, eventually the value of $x_n$ will surpass $L$. But we assumed that $L$ was an upper bound of the sequence, a contradiction. Hence, the sequence is unbounded. It is easy to show that the sequence is increasing. Hence the sequence is increasing and unbounded and therefore must tend to $\infty$.

Asier Calbet
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