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October 25th, 2018, 09:00 PM   #1
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Monotone Convergence Theorem and Cauchy Convergence Criterion

Hello.

I would be grateful for anyone who can solve this problem for me :

Show that the sequence $\left(z_n\right)$ given by $z_{n+1}=\sqrt{a+z_n}$ for $a>0, z_1>0$ converges by using Monotone Convergence Theorem (or Cauchy Convergence Criterion)!
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October 25th, 2018, 10:14 PM   #2
SDK
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Here is a hint: Apply the mean value theorem to successive terms

\[|z_{n+1} - z_{n}| = |\sqrt{a + z_n} - \sqrt{a + z_{n-1}}| \leq \frac{ |z_n - z_{n+1}|}{2\sqrt{a}} \]
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October 25th, 2018, 10:15 PM   #3
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$$\left.\begin{aligned}z_{n+1} &= \sqrt{a+z_n} \\ z_1 &\gt 0 \end{aligned} \right\} \implies z_{n} \gt 0 \; \forall n \ge 1$$
This means that we can compare the magnitude of $z_{n+1}^2$ and $z_n^2$ as proxies for the magnitude of $z_{n+1}$ and $z_n$.

\begin{align} z_{n+1}^2 &= a+z_n \\ z_{n+1}^2 - z_n^2 &= a + z_n - z_n^2 \end{align}
The quadratic expression on the right factorises as $\big(b - z_n\big)\big((b-1) + z_n\big)$ where $b(b-1)=a$ which has a unique solution with $b \gt 1$ because $a \gt 0$.

Thus we have three cases:
  1. $z_n < b$: in which case $\big(b - z_n\big)\big((b-1) + z_n\big) \gt 0$ and thus $z_{n+1}^2 > z_n^2 \implies z_{n+1} > z_n$
  2. $z_n > b$: in which case $\big(b - z_n\big)\big((b-1) + z_n\big) \lt 0$ and thus $z_{n+1}^2 < z_n^2 \implies z_{n+1} < z_n$
  3. $z_n = b$: in which case $\big(b - z_n\big)\big((b-1) + z_n\big) = 0$ and thus $z_{n+1}^2 = z_n^2 \implies z_{n+1} = z_n$

Now, we compare the magnitude of $z_{n+1}^2$ and $b^2$ as proxies for $z_{n+1}$ and $b$.

\begin{align}z_{n+1}^2 - b^2 &= a + z_n - b^2 \\ &= b(b-1) + z_n -b^2 \\ &= z_n -b\end{align}
Here,
  1. if $z_n < b$ we have $z_{n+1}^2 < b^2$ and so $z_{n+1} < b$ too.
  2. if $z_n > b$ we have $z_{n+1}^2 > b^2$ and so $z_{n+1} > b$ too.
  3. and if $z_n = b$ we have $z_{n+1}^2 = b^2$ and so $z_{n+1} = b$ too.

Therefore:
  1. If $z_n < b$ we have $z_n < z_{n+1} < b$
  2. If $z_n > b$ we have $z_n > z_{n+1} > b$
  3. If $z_n = b$ we have $z_n = z_{n+1} = b$

Thus:
  1. If $z_1 < b$, the sequence is increasing and bounded above by $b$.
  2. If $z_1 > b$, the sequence is decreasing and bounded below by $b$.
  3. If $z_1 = b$, the sequence is constant.
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Last edited by v8archie; October 25th, 2018 at 10:18 PM.
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October 25th, 2018, 11:06 PM   #4
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Thanks for such a great hint, SDK. I don't think I could use the hint for now, but of course it would be useful in my next course of advanced real analysis.
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October 25th, 2018, 11:09 PM   #5
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Thank you very much, v8archie. This is what I am looking for.
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