My Math Forum Examine the convergence of a sequence $\{a_{n}\}$ which is given by $a_{1}=a>0,a_{2}=  User Name Remember Me? Password  Calculus Calculus Math Forum  June 6th, 2015, 05:33 AM #1 Newbie Joined: Jun 2015 From: On the Moon. Posts: 1 Thanks: 0 Examine the convergence of a sequence$\{a_{n}\}$which is given by$a_{1}=a>0,a_{2}= I used inequality between arithmetic and geometric means to show that a sequence $\{a_{n}\}$ is bounded: $$a_{n+2}=\sqrt{a_{n+1}a_{n}}\le \frac{a_{n+1}+a_{n}}{2}$$ Solving this, I get quadratic inequality $$a^2_{n+1}-2a_{n+1}a_{n}+a^2_{n}\ge 0$$ which gets me to $$0 \le a_{n}\le 1$$ thus, sequence is bounded. Is this right? How to show that the sequence is monotonic? We don't know if $$a_{n+2}>a_{n+1}$$ or a_{n+2}
 June 6th, 2015, 06:43 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra It's not monotonic because $a \le \sqrt{ab} \le b$ when $a \le b$, but the subsequences $\{\ldots, a_{n-2},a_n,a_{n+2},\ldots\}$ are monotonic. You should concentrate on proving that both subsequences converge to the same value. You can probably guess the value the converge to, and this can be useful.
 June 7th, 2015, 09:14 PM #3 Newbie   Joined: Aug 2014 From: Kalispell MT USA Posts: 20 Thanks: 0 An can = 2 when 0=A over N=√ of 4 But greater then 1 is N over 0

 Tags $a1a>0,$an\$, calculus, convergence, examine, sequence

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Chemist@ Calculus 7 November 1st, 2014 02:16 PM PetraLenski Calculus 3 November 2nd, 2011 03:02 PM elimax Real Analysis 3 July 1st, 2011 09:21 AM fermatprime371 Real Analysis 1 January 30th, 2009 09:13 AM poincare4223 Real Analysis 1 January 14th, 2009 04:59 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top