limit of Recurrence relation

Question

Let $c>0$. Let a sequence $(a_n)$:

$a_1=1$, $a_{n+1}=\frac{1}{2}(a_{n}+\frac{c}{a_{n}})$.

Calculate $$\lim_{n \mapsto\infty }a_n$$

So, my solution is: I can show in induction that $a_n$ is determinated for all $n$. I showed in induction that $(a_n)$ is trupped between $1$ and $c$, (by seperating the cases of $c>1$ and $c<1$).

I guess I need to separate for the 2 above cases and show that the sequence is monotonous and then I can conclude that the sequence converges, and show that the limit is $\sqrt c$. But I cannot know for sure that there's exist a limit to the sequence.

This a recurrent question (no intended pun) about what is known as the babylonian algorithm: see (https://en.wikipedia.org/wiki/Methods_of_computing_square_roots). — Jean Marie, Aug 26 '17 at 19:53
And another version https://math.stackexchange.com/questions/2384298/find-the-limit-if-it-exists-of-s-n1-frac12s-n-fracas-n/ — rtybase, Aug 26 '17 at 22:22

score 0 · Answer 1 · answered Aug 26 '17 at 18:45

0

$$a_{n+1}-\sqrt{c}=\frac{(a_n-\sqrt{c})^2}{2a_n}\geq0$$ and $$a_{n+1}-a_n=\frac{c}{2a_n}-\frac{a_n}{2}=\frac{(\sqrt{c}-a_n)(\sqrt{c}+a_n)}{2a_n}\leq0.$$ Thus, $a_n\rightarrow\sqrt{c}$.

answered Aug 26 '17 at 18:45

Michael Rozenberg

194,933

one must prove that $$a_n>0$$ for all $n>0$ – Dr. Sonnhard Graubner Aug 26 '17 at 18:50
Sonnhard, It's obvious. See the definition of the sequence. – Michael Rozenberg Aug 26 '17 at 18:52
ok then it is obvious, my name is Sonnhard. – Dr. Sonnhard Graubner Aug 26 '17 at 18:53
@ Math Lover I proved that our sequence decreases and $a_{n+1}\geq\sqrt{c}$, which says that there is a limit. – Michael Rozenberg Aug 26 '17 at 18:57
@Math Lover Yes, of course, for $n\geq2$. – Michael Rozenberg Aug 26 '17 at 19:02
@Math Lover See please my previous comment. – Michael Rozenberg Aug 26 '17 at 19:06
@MichaelRozenberg Thanks. Got it! – Math Lover Aug 26 '17 at 19:08
@Math Lover You are welcome! – Michael Rozenberg Aug 26 '17 at 19:09

limit of Recurrence relation

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