Convergence of a Recursive Sequence

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Homework Statement



A sequence {an} defined recursively by a1=1 and an+1=\frac{1}{2+a subn}, n\geq1. Show that the sequence is convergent.

Homework Equations


If a sequence is bdd below and decreasing or it is bdd above and increasing, then it is convergent.

The Attempt at a Solution


{an}\geq0, hence it is bounded below. I checked a few terms from the beginning and obtained a decreasing sequence. I tried induction to show this, but it didn't work. Base case is O.K. a0>a1 Suppose ak>ak+1.
Then we get
ak+2=\frac{1}{2+a sub(k+1)}>\frac{1}{2+a subk}=ak+1, and ak+2>ak+1. I feel like I'm doing a stupid mistake, and I can't understand, if this is true, why the induction does not work. I would be grateful if you could help me. Thanks for your time and effort :)
 
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Induction doesn't always work since the sequence is not descreasing. If you check more terms then you see that.

Are you familiar with the contraction mapping theorem by Banach?
 
micromass said:
Induction doesn't always work since the sequence is not descreasing. If you check more terms then you see that.

Are you familiar with the contraction mapping theorem by Banach?

Ah, really? So that was my mistake... I have no idea about contraction mapping theorem.
Thanks for enlightening me :biggrin:
Is there a more practical way to show a sequence is convergent? :confused:
 
Can I use the limit definition? I see no way out... :(
 
You may be able to show that the "odd" elements of the sequence form a decreasing subsequence and the "even" elements an increasing subsequence ...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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