Convergence of a sequence

**fishfan** · April 3rd 2011, 05:01 AM

Hi everyone,

I have got a question whereby a(n) = 1/(1+a(n-1)). This is a sequence and I am asked to find whether it converges.

I have thought of one way to solve it whereby we assume that it converges e.g to a value "a", and therefore you get a quadratic equation a^2+a=1 and therefore you get a value of a, and therefore we conclude that it converges.

However, it has been bugging me that we have used an assumption to derive the answer. By coming up with the answer after using the assumption, does it not just prove that we have a solution in the case that it converges, and not that it actually converges per se? If it actually diverges for some parameters, e.g. when a(n) belongs to some domain, then won't we be missing this part of the solution?

Although this may be a handful to read, I would really appreciate if anyone can enlighten me on this.

Thank you!

**Sambit** · April 3rd 2011, 05:16 AM

Originally Posted by fishfan

I have thought of one way to solve it whereby we assume that it converges e.g to a value "a", and therefore you get a quadratic equation a^2+a=1 and therefore you get a value of a, and therefore we conclude that it converges.

You can do this only if you know it converges. And to prove convergence, you will have to prove that the function is monotonically increasing (or decreasing) and bounded above (or below).

**chisigma** · April 3rd 2011, 06:03 AM

The 'recursive relation' can be written as...

$\displaystyle \Delta_{n}= a_{n+1}- a_{n}= \frac{1}{1+a_{n}} - a_{n} = f(a_{n})$ (1)

For semplicity sake we analyse the case $a_{0}>-1$ . The $f(x)$ for $x>-1$ is illustrated here...

There is only one 'attractive fixed point' in $x_{0}= .618039887...$ and because in the entire range of definition is $|f(x)|< |2\ (x_{0}-x)|$ [red line...] but also $|f(x)|>|x_{0}-x|$ [blue line...] , any $a_{0}>-1$ will generate a sequence converging at $x_{0}$ 'oscillating' ...

... that means that we have a convergent sequence that is neither 'monotonically increasing' nor 'monotonically decreasing'...

Kind regards

$\chi$ $\sigma$

**fishfan** · April 3rd 2011, 06:42 AM

Wow thanks for the diagram and everything chisigma. That's really useful! I sort of understand the concept, but not the maths. How do you get the equations for the red and blue line?

**chisigma** · April 3rd 2011, 07:28 AM

Let's suppose to have a recursive equation of first order...

$a_{n+1} = \varphi(a_{n})$ (1)

... where and we know the value $a_{0}$ . If we define...

$\Delta_{n}= a_{n+1} - a_{n} = \varphi(a_{n})- a_{n} = f(a_{n})$ (2)

... the sequence defined by (1) can be written as...

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

The sequence $a_{n}$ converges if and only if converges the series in (3) and necessary condition is that...

$\displaystyle \lim_{n \rightarrow \infty} \Delta_{n}=0$ (4)

Consequence of (4) is that necessary condition for the existence of a finite limit $x_{0}$ is that...

a) $f(x)$ is continous 'somewhere around' $x_{0}$ ...

b) $f(x_{0})=0$ ...

c) $f^{'} (x_{0}) <0$ ...

A point $x_{0}$ which satisfies the condition a), b) and c) is called 'attractive fixed point'. Sufficient condition for having a sequence convergent at $x_{0}$ is that 'somewhere around' $x_{0}$ is...

$|f(x)|< |2\ (x_{0}-x)|$ (5)

In that case we can have two distinct cases...

i) $|f(x)|< |x_{0}-x|$ (6)

... and in that case the we have a 'monotonic series' that converges because for n 'large enough' is...

$\displaystyle |\frac{\Delta_{n+1}}{\Delta_{n}}|<1$ (7)

ii) $|x_{0}-x| < |f(x)|< |2\ (x_{0}-x)|$ (8)

... and in that case we have an 'alternating series' that converges for the 'Leibnitz's criterion'...

Kind regards

$\chi$ $\sigma$

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