# Limit

• Apr 19th 2011, 07:47 AM
unlimited
Limit
Let $a_{1}=\frac{5}{2}$ and $a_{n+1}=\sqrt{ a_{n}^{3} -12a_{n}+\frac{20n+21}{n+1}}$
Prove that the sequence $(a_{n})$ converges and find its limit
• Apr 19th 2011, 09:29 AM
chisigma
Quote:

Originally Posted by unlimited
Let $a_{1}=\frac{5}{2}$ and $a_{n+1}=\sqrt{ a_{n}^{3} -12a_{n}+\frac{20n+21}{n+1}}$
Prove that the sequence $(a_{n})$ converges and find its limit

If the 'recursive relation' is...

... it can be written as...

http://quicklatex.com/cache3/ql_b7fc...5a37e84_l3.png (2)

Now for n 'large enough' is with good approximation...

http://quicklatex.com/cache3/ql_c148...6a5b6fe_l3.png (3)

... and f(x) is represented here...

http://digilander.libero.it/luposabatini/MHF114.bmp

The f(x) has an 'attractive fixed point' at $x_{0}=2$ and a 'repulsive fixed point' at http://quicklatex.com/cache3/ql_0f37...699a5a9_l3.png so that for http://quicklatex.com/cache3/ql_f033...daec15b_l3.png the sequence will converge monotonically at 2...

Kind regards

$\chi$ $\sigma$
• Apr 19th 2011, 12:33 PM
HallsofIvy
A different approach:
IF that sequence converges, then, calling the limit L, we must have
lim a_{n+1}= \SQRT{lim(a_n^3)- 12lim(a_n)+ lim((20n+ 1)/(n+1))}
which gives
L= SQRT{L^3- 12L+ 20}. Squaring both sides, L^2= L^3- 12L+ 20 or L^3- L^2- 12L+ 20= 0. It is easy to see that L= 2 is a root of that. Factoring out (L- 2) leaves L^2+ L- 10= 0 which has roots L= (1+ SQRT{41})/2 and L= (1- SQRT{41})/2.

Here a_1= 5/2 and then a_2= \sqrt{125/4- 30+ 1}= sqrt{9/4}= 3/2< 5/2. It's not too difficult to show that this is a decreasing sequence and all terms are positive. That is, the sequence is decreasing and bounded below (by 0). By the "monotone sequence theorem", it must converge so our first paragraph is valid. Since it starts at 5/2 and is decreasing, that limit must not be (1+ sqrt(41))/2 which is larger than 5/2. Since all terms of the sequence are positive, it cannot converge to (1- sqrt(41))/2. Therefore it must converge to 1.