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Suppose that a sequence $y_n$ is defined iteratively by $y_0 = 1$ and $$ y_{ n + 1 } = \frac{ 1 }{ 2 + y_n } $$ Show that $\{ y_n \}_{ n \geq 0 }$ is a convergent sequence.

I'm not really sure where to start.

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5 Answers 5

Where I would start is to compute a few terms. A spreadsheet makes this easy: put 1 in A1, =1/(2+A1) in B1 and copy down. It converges quickly to $\sqrt 2-1$

If there is a limit $L$, we must have $L=\frac 1{2+L}$, which has solutions $-1\pm \sqrt 2$

Now let $z_n=y_n-(\sqrt 2 -1)$ You should be able to show that $z_n$ is monotonically decreasing.

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Try showing that the sequence is monotonically increasing (after a point), and that all but the starting term of the sequence is bounded above by $\frac12$.

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One strategy is

1)take the limit in the both sides

2)use the the operation rules of the limits

3)Remember $\lim y_{n+1}=\lim y_n$

Then you can have the limit of the sequence.

Note: this doesn't prove that this sequence is convergent.

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@DonAntonio Thank you, it's true –  user42912 Dec 17 '12 at 4:25

Consider trying to prove that two subsequences that together make up the entire function ($y_n$ for each of odd and even $n$) both converge to the same point.

If we try defining $y_{n+2}$ in terms of $y_n$ rather than $y_{n+1}$, we get: $y_{n+2} = \cfrac{1}{2 + \cfrac{1}{2 + y_n}}$ which is $\cfrac{1}{\cfrac{5 + 2y_n}{2 + y_n}}$ or $\cfrac{2 + y_n}{5 + 2y_n}$.

Then, show that if $y_n < \sqrt{2} - 1$, then $y_n < y_{n+2} < \sqrt{2} - 1$, and if $y_n > \sqrt{2} - 1$, then $y_n > y_{n+2} > \sqrt{2} - 1$. Then you'll have proved the result by induction.

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Actually, I'm not sure myself if this works in all cases. It seems kinda sketchy. Does anyone have a counterexample or a clearer explanation of this? –  Joe Z. Dec 17 '12 at 4:36

First prove that $y_n \geq 0$ for all $n$.

Let $L>0$ satisfy $L^2 + 2L = 1$. Then \begin{align*} \left \vert y_{n+1} - L \right \vert & = \left \vert \dfrac1{2+y_n} - L \right \vert = \left \vert \dfrac{1-2L-Ly_n}{2+y_n} \right \vert = \left \vert \dfrac{1-2L-L^2 - Ly_n + L^2}{2+y_n} \right \vert\\ & = \left \vert \dfrac{L (L-y_n)}{2+y_n} \right \vert \end{align*} Note that $L = \sqrt{2} - 1 < \dfrac12$ and $2+y_n > 2$. Hence, $$\left \vert y_{n+1} - L \right \vert \leq \dfrac{\left \vert y_n - L \right \vert}4$$ Hence, $$\left \vert y_{n+1} - L \right \vert \leq \dfrac{\left \vert y_0 - L \right \vert}{4^{n+1}} \to 0$$ Hence, the limit exists and the limit is $L= \sqrt{2} - 1$.

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