Let ($c_n$) be the recursively defined sequence
$c_1$ = 10,
$$c_n = \frac{1}{2}(c_{n-1} + \frac{10}{C_{n-1}} )$$ for all $n$ $\geq$ 2.
Find $$\lim_{x\to\infty} c_n$$ and prove your claim.
So I'm aware I need to be proving $\forall n \in $ $\mathbb{N}$, $c_n \geq \sqrt{10}$ but I m having trouble moving on from here. I'm pretty sure I could figure it out if could work out the boundedness of the sequence. I'm thinking of working with squaring both sides and getting $$(c_n)^2 \geq 10$$
On
First we can see that:$\ c1=10>0 \ We \ assume \ cn>0 \ and \ with \ use \ of \ mathematical \ induction \ easy \ we \ can \ find \ cn>0 \ for \ every \ n. \ From \ a \ known \ theormem \ we \ can \ say \ for \ two(positive) \ numbers\ \frac 12(k+m)>={\sqrt k \sqrt m} .So \ we \ have \left(cn+\frac {10} c_n \right)>{\sqrt cn \sqrt \frac {10} c_n=\sqrt 10>0} .So \ we\ proved \ that \ the \ sequnce \ is \ blocked \ down\ .If \ we \ take\ the\ difference \ c_{n+1}-cn=\frac{1}{2}(c{n} + \frac{10}{C_{n}} )-cn=\frac{1}{2c_n}(10-cn^2){}<=0 .So \ we\ proved \ the \ sequence \ is \ also:\ descending \ sequence.Now \ from \ a \ well \ known \ theorem \ "Bolzano–Weierstrass" \ the \ above \ sequence \ is \ convergent \ sequence.\ So \ it's \ easy \ now \ if \ we \ say \ the \ limit \ of \ the \ sequence \ is \ x \ to \ solve \ the \ equation \ x=\frac 12 \left(x+\frac {10} x \right) \ and \ finally \ x=\sqrt10 $
If the limit does exist, it is the solution of $$x=\frac 12 \left(x+\frac {10} x \right)\implies x^2=10\implies x= ???$$ Otherwise, if you rewrite $$c_{n+1}=c_n-\frac{c_n^2-10}{2c_n}$$ you can recognize Newton iteration scheme for finding the zero of equation $$f(c)=c^2-10$$ starting at $c_0$.