1. ## 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

2. 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...

(1)

... it can be written as...

(2)

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

(3)

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

The f(x) has an 'attractive fixed point' at $x_{0}=2$ and a 'repulsive fixed point' at so that for the sequence will converge monotonically at 2...

Kind regards

$\chi$ $\sigma$

3. 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.