I have a sequence:
an=√3+√3+...√3 , it repeats n-times.
and i have to prove that it is a Cauchy's sequence.
So i did this:
As one theorem says that every convergent sequence is also Cauchy, so i proved that it's bounded between √3 and 3 (with this one i am not sure, please check if i am right with this one.)And also i proved tat this sequence is monotonic. (with induction i proved this: an≤an+1
so if it's bounded and monotonic, therefore it is convergent and Cauchy.
I am just wondering if this already proved it or not? And also if the upper boundary - supremum if you wish - is chosen correctly.
I appreciate all the help i get.
Answer
an+1=√an+3 ⇒a2n+1=an+3 as n→∞ ⇒a2n+1=an+3 x2=x+3⇒ x2−x−3=0 andit convergents to the x=1+√132
No comments:
Post a Comment