Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

5 Answers

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.

share|improve this answer
add comment

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 $\frac12$.

share|improve this answer
add comment

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.

share|improve this answer
    
@DonAntonio Thank you, it's true –  user42912 Dec 17 '12 at 4:25
add comment

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.

share|improve this answer
    
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
add comment

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 $L= \sqrt{2} - 1$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.