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I would like to show that the sequence $(x_n)$ defined by $x_{n+1}=x_n^2+\frac{1}{4},x_0=0$ is convergent. To do this it suffices to show that the sequence is bounded and monotonic. If the sequence converges, I know it must be bounded above by $1$. Otherwise, let $x_n>1$ for some $n$. Then for each $x_{n+1}$, $x_{n+1}>x_n$, since $x^2>x$ for all $x>1$. This would imply that the sequence is divergent if it is not bounded above by $1$.

To show monotonicity, consider the following: $$x_{n+1}-x_n=x_n^2-x_n-\frac{1}{4}=0\implies x_n=\frac{1\pm\sqrt 2}{2}$$ These roots occur when $x_n<0$ and $x_n>1$, but the sequence $(x_n)$ is bounded above by $1$ and below by $0$, so $x_{n+1}-x_n$ is either strictly greater than or less than $0$. It is easy to check that $x_1>x_0$, so the sequence is increasing and therefore monotone.

I would like to verify that the work I have so far is correct. From here we have a hint that says to check that any limit $x$ must satisfy $x=x^2+\frac{1}{4}$ to show that the sequence converges to $\frac{1}{2}$.

I am not convinced that what I have for boundedness is correct, as it relies on the fact that the sequence is convergent when this is what we are trying to prove.

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    $\begingroup$ Wether $(x_n)$ converges or not clearly depends on $x_0$. $\endgroup$ – Olivier Roche Sep 6 '19 at 9:27
  • $\begingroup$ @5xum sorry, I forgot to add that $x_0=0$. $\endgroup$ – csch2 Sep 6 '19 at 9:27
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    $\begingroup$ Try proving the induction statement "for all $n\ge 0$, $x_n<\frac{1}{2}$ and $x_{n+1}>x_n$" $\endgroup$ – Paul Sep 6 '19 at 9:35
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If the sequence converges, I know it must be bounded above by $1$

How do you know that?


for each $x_{n+1}$, $x_{n+1}>x_n$, since $x^2>x$ for all $x>1$. This would imply that the sequence is divergent if it is not bounded above by $1$.

There are many convergent sequences for which $x_{n+1}>x_n$, so your logic here is flawed.


Your entire proof that the sequence is bounded above by $1$ is not a proof at all. It is not clear at all what the logical path leading from initial assumptions to conclusions is. You start the proof saying "if the sequence is convergent". You cannot just assume the sequence is convergent!


To show monotonicity, consider the following: $$x_{n+1}-x_n=x_n^2-x_n-\frac{1}{4}=0\implies x_n=\frac{1\pm\sqrt 2}{2}$$ These roots occur when $x_n<0$ and $x_n>1$, but the sequence $(x_n)$ is bounded above by $1$ and below by $0$, so $x_{n+1}-x_n$ is either strictly greater than or less than $0$.

This is very confusingly written, and also wrong. $x_{n+1}-x_n=x_n^2-x_n+\frac14.

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    $\begingroup$ The expression for $x_{n+1}-x_n$ is incorrect. $\endgroup$ – Peter Foreman Sep 6 '19 at 9:40
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We have $x_0=0$

  1. Show by induction: $0 \le x_n \le 1/2$ for all $n$.

  2. $x_{n+1}-x_n=x_n^2-x_n+\frac{1}{4}=(x_n-\frac{1}{2})^2 \ge 0$ for all $n$.

(you wrote $x_{n+1}-x_n=x_n^2-x_n-\frac{1}{4}$, which is false)

Can you proceed ?

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Hint

I think you mad a typo. Actually$$x_{n+1}-x_n=x_n^2-x_n+\frac{1}{4}=\left(x_n-{1\over 2}\right)^2\ge 0$$To show the boundedness, from $0\le x_n<{1\over 2}$ conclude that $0\le x_{n+1}<{1\over 2}$.

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  • $\begingroup$ Thank you, I did make a typo. I have figured out the boundedness now by induction. $\endgroup$ – csch2 Sep 6 '19 at 9:42
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    $\begingroup$ You're welcome. Have fun with induction! $\endgroup$ – Mostafa Ayaz Sep 6 '19 at 9:43

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