Letting $a_{1}=2$ I have a recursive sequence defined as follows: $$ a_{n+1} = \frac{a_{n}}{2} + \frac{5}{a_{n}} \ \ \forall n \geq1$$
How can I prove that the sequence {$a_{n}$} converges and also find the limit of this sequence? I plugged in first few terms and get something like:
2,
3.5,
3.179,
3.162,
3.162,
...
The textbook example showed an example of a recursive sequence that was monotone, but this sequence I have here does not appear as such. However, this sequence looks like a Cauchy sequence though. I am completely lost and not sure how to proceed. Any help is much appreciated! Thank you.
By induction we see that $a_n>0$ and since $$a_{n+1}-\sqrt{10}=\frac{(a_n-\sqrt{10})^2}{2a_n}\geq0,$$ by induction we obtain $$a_n>\sqrt{10}$$ for all $n\geq2.$
In another hand, $$a_{n+1}-a_n=\frac{5}{a_n}-\frac{a_n}{2}=\frac{(\sqrt{10}-a_n)(\sqrt{10}+a_n)}{2a_n}<0,$$ for all $n\geq2,$ which gives that $a$ decreases.
Id est, there is $\lim\limits_{n\rightarrow+\infty}a_n$ and let this limit is equal to $a$.
We obtain $$a=\frac{5}{a}+\frac{a}{2},$$ which gives $a=\sqrt{10}.$