# Prob. 7, Sec. 3.3, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: Is this sequence convergent?

Here is Prob. 7, Sec. 3.3, in the book Introduction to Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:

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

My Attempt:

We note that $x_1 = a > 0$. Suppose that $x_k > 0$ for some $k \in \mathbb{N}$. Then $$x_{k+1} = x_k + \frac{1}{x_k} > 0.$$ Therefore $$x_n > 0 \ \mbox{ for all } n \in \mathbb{N}.$$

We note that for any natural number $n$, as $x_n > 0$, so we have $$x_{n+1} = x_n + \frac{1}{x_n} = \frac{1}{2} \left[ \left( \sqrt{x_n} + \frac{1}{\sqrt{x_n} } \right)^2 + \left( \sqrt{x_n} - \frac{1}{\sqrt{x_n}} \right)^2 \right] \geq \frac{1}{2} \left[ \left( \sqrt{x_n} + \frac{1}{\sqrt{x_n} } \right)^2 - \left( \sqrt{x_n} - \frac{1}{\sqrt{x_n}} \right)^2 \right] = 2.$$ Thus we can conclude that $$x_n \geq 2 \ \mbox{ for all } n \in \mathbb{N} \ \mbox{ such that } n \geq 2. \tag{0}$$

Now suppose $\left( x_n \right)$ is convergent, with limit $x$ in $\mathbb{R}$. Then as $x_n \geq 2$ for all $n \in \mathbb{N}$ and $n \geq 2$, so we must have $x \geq 2$. Then we must have $$x = x + \frac{1}{x},$$ which implies that $$\frac{1}{x} = 0,$$ which is impossible.

Hence our sequence cannot converge in $\mathbb{R}$.

For each $n \in \mathbb{N}$, as $x_n > 0$, so we have $$x_{n+1} = x_n + \frac{1}{x_n} > x_n.$$ Thus our sequence is strictly increasing.

As our sequence is monotonically increasing and not convergent, so it is not bounded from above. Therefore we can conclude that $$\lim_{n \to \infty} x_n = +\infty.$$

Is what I have done so far correct and accurate? If not, then where are the problems?

• No problem with what you wrote. The part showing that $x_n\ge2$ does not seem necessary to reach the conclusion. – Did Mar 23 '18 at 11:02
• Why has my post been down-voted, I wonder? What've I done wrong? – Saaqib Mahmood Apr 3 '18 at 17:30

You can prove that the sequence is monotonically increasing by looking at the ratio $\frac{x_{n+1}}{x_{n}}$.
$$\frac{x_{n+1}}{x_{n}} =1 + \frac{1}{x_{n}^{2}} > 1$$ and we know that $x_{1} > 0$, so we must have the sequence $x_{n}$ strictly increasing.
Or another way, use induction : $x_{1} > 0$, then assume $x_{k} > 0, \:\: k \in \mathbb{N}$, then $$x_{k+1} = 1 + \frac{1}{x_{k}} > 0$$ so by induction, we conclude that $x_{n} > 0$. Then notice that $$x_{n+1}-x_{n} = \frac{1}{x_{n}} >0$$ so the sequence is strictly increasing.