Limit of ratio of nested radicals

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SUMMARY

The sequence defined by the recurrence relation an+1 = √(2 + an) with a0 = √2 converges to 2 as n approaches infinity. Additionally, the limit lim {(an+1 - 2) / (an - 2)} equals 1/4. The solution involves factorizing (an+12 - 4) and comparing it with the derived expression from the recurrence relation.

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garrus
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Homework Statement


I have a sequence an+1=sqrt(2+an) ,with a0 = sqrt(2)
which leades to nested radicals.

I am asked to show that for n approaching infinity:
1) the sequence converges to 2, and that

2) lim {(an+1 -2) / (an -2)} = 1/4


The Attempt at a Solution


1)I have proven the convergence to 2.
2) for the 2nd limit, i tried de L'Hospital's rule, since we have 0 / 0 , but can't figure out exactly how to find the derivative of an (infinite) nested roots.
- i tried also square differences , but still get to 0/0.

Any hints?
 
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garrus said:
Any hints?

Yep. Factorize (a_{n+1}^2 - 4) and compare with the same expression derived from your recurrence relation.
 
Got it.
Thanks for your help, should've noticed that =[
 

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