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Is it possible to prove that the sequence $(a_n)_{n\ge1}$ defined as $$a_{n+1}=\frac{1}{2}\left(a_n+\frac{4}{a_n}\right)$$ convergent without the condition $a_1>2$?

Context: Yesterday, this problem was asked on an entrance exam and I really struggled to find a solution for this. Later, I remembered that this is actually baby Rudin's exercise $16(a)$ chapter $3$. Moreover, even if it is possible to somehow prove the convergence, the question also asked to find the limit and without this condition on the first term it is impossible to decide if the limit is actually $2$ or $-2$. So, before informing the university, if somebody could also confirm my observation I would really appreciate it.

bob
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Anish Ray
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1 Answers1

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I provide another solution than those in the link. With this solution, we can have closed-form formula for $a_n$.

If $a_1$ is negative then all $(a_n)_n$ are negative, and it suffices to denote $b_n= -a_n$.

So, we can suppose that $a_1$ are positive and so are all $(a_n)_n$. We notice that $$a_{n+1}-2 = \frac{1}{2}\left(\frac{a_n-2}{\sqrt{a_n}} \right)^2 \tag{1}$$ $$a_{n+1}+2 = \frac{1}{2}\left(\frac{a_n+2}{\sqrt{a_n}} \right)^2\tag{2}$$

From $(1),(2)$ we have

$$\frac{a_{n+1}-2 }{a_{n+1}+2 }= \left(\frac{a_n-2}{a_n+2} \right)^2=...=\left(\frac{a_1-2}{a_1+2} \right)^{2^n} \tag{3}$$

From $(3)$ you deduce the closed form solution of $a_n$.

From that, it's easy to show that, $a_n \to 2$ when $n\to +\infty$.

Return back to the case where $a_1 <0$ at the beginning, we can prove that $a_n = - b_n\to -2$ when $n\to +\infty$

Arctic Char
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NN2
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