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

We have $x_1= a > 0, x_2= a+\dfrac{1}{a}, (\dfrac{1}{a} >0)$ which implies that $x_2 > x_1$

Then assume for induction: $x_k < x_{k+1} < a+1 $ (since $\dfrac{1}{x_k} < 1$)

$$x_k + \frac{1}{x_k} < x_{k+1}+ \frac{1}{x_{k+1}} < a + 1$$

$$\implies x_{k+1} < x_{k+2} < a + 1$$

This shows that that $x_k < x_{k+1} \Rightarrow x_{k+1} < x_{k+2}$. Therefore $x_n < x_{n+1}$ for all $n \in \mathbb N$.

It follows by the monotone convergence theorem that $x_n$ is increasing and bounded.

StubbornAtom
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axanalysis
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2 Answers2

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hint: $x_n^2 \geq x_{n-1}^2 + 2 \Rightarrow x_n^2 \geq 2(n-1) \Rightarrow x_n \geq \sqrt{2(n-1)}$

DeepSea
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$(x_n)$ is monotonically increasing is clear. Let's prove that it is unbounded. \begin{equation} x_{n+1}=x_n+\frac{1}{x_n}\\ x_{n+1}^2=\left(x_n+\frac{1}{x_n} \right) ^2\\ x_{n+1}^2=x_n^2+\frac{1}{x_n^2}+2\\ \geq x_n^2+2 \end{equation} $\therefore x_n^2 \geq ...\geq a^2+2(n-1) \Rightarrow x_n\geq \sqrt{a^2+2(n-1)}$. It can be clearly seen that the right hand side of the equation is unbounded and hence the left hand side.

Chandramauli Chakraborty
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