cauchy convergence

**Webby** · September 5th 2010, 05:58 AM

Hi I am having a bit of trouble with a couple of questions in regards to convergences.
Let the sequence $\{a_{n}\}^\infty _{1}$

let $a_{1} =1, a_{2} = 2, a_{n+2} = \frac{a_{n+1} + a_{n}}{2}$

Prove that $|a_{n+2} - a_{n+1}| = \frac {1}{2^n}$

Hence prove that $\{a_{n}\}^\infty_{1}$

From my attempts at part 1 I have been able to show that for n=0 and n=1 that the function will be equal to $\frac{1}{2^0}$ and $\frac{1}{2^1}$ respectively. However I am having trouble actually proving for n terms that it is true. I am able to get to $|\frac{a_{n+1} - a_{n-1}}{2}|$ but am completely stuck there on.

For the second part I have very little idea as to how I can prove that the sequence converges using my result from part 1.
Thanks you for any input in advance :d

**alexmahone** · September 5th 2010, 07:55 AM

Use induction for part 1.

$|a_{m+2}-a_{m+1}|=|\frac{a_{m+1}+a_m}{2}-a_{m+1}|$

= $|\frac{a_m-a_{m+1}}{2}|$

Thus, if $|a_{m+1}-a_{m}|=\frac{1}{2^{m-1}}$ ,

$|a_{m+1}-a_{m}|=\frac{1}{2*2^{m-1}}=\frac{1}{2^m}$

**chisigma** · September 5th 2010, 08:08 AM

Setting $\alpha_{n} = a_{n+1} - a_{n}$ the difference equation becomes...

$\displaystyle \alpha_{n+1}= -\frac{\alpha_{n}}{2}$ , $\alpha_{0}=1$ (1)

... and its solution is...

$\alpha_{n} = (-\frac{1}{2})^{n}$ (2)

The sequence $a_{n}$ is given by...

$\displaystyle a_{n}= a_{0} + \sum_{k=0}^{n} \alpha_{n} = 1+\frac{2}{3}\ \{1-(-\frac{1}{2})^{n}\}$ (3)

... and from (3) You derive that is...

$\displaystyle \lim_{n \rightarrow \infty} a_{n} = \frac{5}{3}$ (4)

Kind regards

$\chi$ $\sigma$

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