# Recursive sequence - Convergency and Limit

i just started university so im pretty new to all this new math. My problem is to solve this recursive sequence: $a_{n+1} = a_{n}^3$ with: $a_{0} = \frac{1}{2}$ and: $n \in N$

I've to analyse convergence and if its convergent i've to get the limit of this sequence.

I dont know how to start and this to the power of three confuses me.

thanks in advance!

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Hi John, welcome to math.stackexchange! I suggest for your problem to write out the first 4 or 5 terms of the sequence and see if you notice a pattern. Maybe you can guess if it converges or not from that. – Ragib Zaman Apr 11 '12 at 15:10
We have $0<a_n<(1/2)(1/2)^n$. – André Nicolas Apr 11 '12 at 16:20

## 3 Answers

Due to $a_{n+1}=a_n^{3}=(a_{n-1}^3)^3=a_{n-1}^{3^2}=……=a_{0}^{3^{n+1}}=(\frac{1}{2})^{3^{n+1}}$. if you know this ,you can do this question.

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When the limit exists we have $\lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} a_{n+1}=a$, so to find the limit you need to solve the equation: $a^3=a$.

But that's not enougth, we can see from the definition of $a_n$ that it's non-negative (why?), and its decreasing to zero, so we can infer from above that limit is zero (a decreasing sequence converges to the infimum of the sequence).

To show that the sequence is decreasing we need to show that $a_n \geq a_{n+1}=a^3_n \ \forall n\in \mathbb{N}$, this happens when $a_n (1-a^2_n)\geq 0$, this is always valid cause $0\leq a_n \leq 1$ (why?).

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BTW, you can prove by induction quite easily that $a_n \in [0,1]$. – MathematicalPhysicist Apr 11 '12 at 15:21

Another idea is to define a new sequence $b_n=\log a_n$. Then your recursive equation $$a_{n+1}=a_n^3$$ becomes $$b_{n+1}=3b_n$$ which is pretty straight forward.

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