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Came across the following exercise in Bartle's Elements of Real Analysis and am a little unsure about my solution. Would be extremely grateful if someone could verify it for me.

Let $x_1 \in \Bbb R$ satisfy $x_1 \gt 1$ and let $x_{n + 1} = 2 - \frac{1}{x_n}$ for $n \in \Bbb N$. Show that the sequence $(x_n)$ is monotone and bounded. What is the limit?

My Attempt:

Part 1 - (Showing monotony and bound)

We will prove that $1 \lt x_{n +1} \lt x_n$ for each $n \in \Bbb N$. $x_2 = 2 - \frac 1 {x_n} \gt 2 - 1 = 1$ and $x_2 = 2 - \frac 1 {x_2} = \dfrac{2x_2 - 1}{x_2} = 1 + \dfrac{x_2 -1}{x_2} \lt 1 + x_2 - 1 = x_2$ whence we have that $1 \lt x_2 \lt x_1$. Now suppose $1 \lt x_{n +1} \lt x_n$ for an arbitrary natural number $n$.

$$x_{n + 2} = 2 - \frac{1}{x_{n+ 1} } \gt 2 - 1 = 1$$

$$ x_{n + 2} = 2 - \frac{1}{x_{n+ 1} } = \frac{2x_{n + 1} - 1}{x_{n + 1}} = 1 + \frac{x_{n + 1} - 1}{x_{n + 1}} \lt 1 + x_{n + 1} - 1 = x_{n + 1} $$

Therefore we have $ 1 \lt x_{n + 2} \lt x_{n + 1} $ and hence induction is complete.

So the sequence is monotone decreasing and is bounded below by $1$. By the Monotone Convergence Theorem $\lim (x_n) = x$ exists in $\Bbb R$.

Part 2 - (Finding Limit)

We know that $x_nx_{n + 1} = 2x_n - 1$ and that $(x_n)$ converges and so does $(x_{n + 1})$ to the same limit since it can be interpreted as a subsequence of $(x_n)$. Then the sequence $(x_nx_{n + 1})$ will also converge to the limit $\lim {(x_n)} \lim {(x_{n+ 1})} = x^2 $. The sequences $(y_n)$ and $(z_n)$ given by $y_n = 2$ and $ z_n = -1 $ for $n \in \Bbb N$ also converge to $2$ and $(-1)$ respectively so $\lim (2x_n - 1) = 2x - 1$. Whence we have that $$x^2 = 2x - 1 \implies x = 1$$


  • Am a little unsure about two areas of the proof - one, the part which uses induction since I found both steps can be shown using the same method and two, the computation of $\lim (x_n)$. So any criticism is more than welcome.
  • If there is an easier way to prove this I would very much like to know since this is just the first exercise and I do not want to follow a wrong path.
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  • $\begingroup$ For the first part I used to do with separate induction arguments. One to show its bounded and the other to show its decreasing. $\endgroup$ – user60887 Mar 16 '14 at 1:30
  • $\begingroup$ @user60887: I don't mind combining the two actually. My doubts stem from the fact that the base case and the induction step are pretty much the same which is, you know, unusual. $\endgroup$ – Ishfaaq Mar 16 '14 at 1:32
  • $\begingroup$ That can definitely happen with some induction proofs. But I feel for your proof to be complete id rather just show it all instead of saying by the method in the base case this follows which is not complete. $\endgroup$ – user60887 Mar 16 '14 at 1:37
  • $\begingroup$ @user60887: Edited in. $\endgroup$ – Ishfaaq Mar 16 '14 at 16:09
  • $\begingroup$ Yes i think the proof looks good to me as well as the calculation of the limit itself. $\endgroup$ – user60887 Mar 16 '14 at 17:19
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To take this out of unanswered list:

We will prove that $1 \lt x_{n +1} \lt x_n$ for each $n \in \Bbb N$. $x_2 = 2 - \frac 1 {x_n} \gt 2 - 1 = 1$ and $x_2 = 2 - \frac 1 {x_2} = \dfrac{2x_2 - 1}{x_2} = 1 + \dfrac{x_2 -1}{x_2} \lt 1 + x_2 - 1 = x_2$ whence we have that $1 \lt x_2 \lt x_1$. Now suppose $1 \lt x_{n +1} \lt x_n$ for an arbitrary natural number $n$.

$$x_{n + 2} = 2 - \frac{1}{x_{n+ 1} } \gt 2 - 1 = 1$$

$$ x_{n + 2} = 2 - \frac{1}{x_{n+ 1} } = \frac{2x_{n + 1} - 1}{x_{n + 1}} = 1 + \frac{x_{n + 1} - 1}{x_{n + 1}} \lt 1 + x_{n + 1} - 1 = x_{n + 1} $$

Therefore we have $ 1 \lt x_{n + 2} \lt x_{n + 1} $ and hence induction is complete.

So the sequence is monotone decreasing and is bounded below by $1$. By the Monotone Convergence Theorem $\lim (x_n) = x$ exists in $\Bbb R$.

We know that $x_nx_{n + 1} = 2x_n - 1$ and that $(x_n)$ converges and so does $(x_{n + 1})$ to the same limit since it can be interpreted as a subsequence of $(x_n)$. Then the sequence $(x_nx_{n + 1})$ will also converge to the limit $\lim {(x_n)} \lim {(x_{n+ 1})} = x^2 $. The sequences $(y_n)$ and $(z_n)$ given by $y_n = 2$ and $ z_n = -1 $ for $n \in \Bbb N$ also converge to $2$ and $(-1)$ respectively so $\lim (2x_n - 1) = 2x - 1$. Whence we have that $$x^2 = 2x - 1 \implies x = 1$$

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