# Convergence of a sequence

• Apr 3rd 2011, 05:01 AM
fishfan
Convergence of a sequence
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!
• Apr 3rd 2011, 05:16 AM
Sambit
Quote:

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).
• Apr 3rd 2011, 06:03 AM
chisigma
The 'recursive relation' can be written as...

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

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

http://digilander.libero.it/luposabatini/MHF113.bmp

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Apr 3rd 2011, 06:42 AM
fishfan
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?
• Apr 3rd 2011, 07:28 AM
chisigma
Let's suppose to have a recursive equation of first order...

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$