x_1=3/2, x_(n+1)=3/(4-x_n)

prove that the series x_n converges and find its limit.

the sequence is

3/2, 6/5, 15/14, ...

I don't see how it can converge, since each term is greater than 1.

I can answer the second part. Obviously it converges to 1.

But i do not know how to prove it...

I do not know if it is legit but maybe since you know the limit is 1 use definition: For every n>N |a_n-1|<epsilon

and prove this way.

its no proof but you could see the results are: 3/2, 6/5, 15/14,42/41,123/122,366/365,1095/1094,3282/3281,9843/9842.

as you could see it comes closer and closer to 1 therfore its limit is 1.

sorry. My bad. I was thinking we wanted the series to converge, not the sequence.

To prove that the series x_n converges and find its limit, we will use the concept of a recursive sequence.

Let's start by finding a pattern in the sequence.

Given x_1 = 3/2, and x_(n+1) = 3/(4 - x_n), we can start generating the terms of the sequence:

x_1 = 3/2
x_2 = 3/(4 - x_1) = 3/(4 - 3/2) = 3/(5/2) = 6/5
x_3 = 3/(4 - x_2) = 3/(4 - 6/5) = 3/(14/5) = 15/14
x_4 = 3/(4 - x_3) = 3/(4 - 15/14) = 3/(41/14) = 42/41
...

We can observe that as n increases, the numerators and denominators of the terms get larger, indicating that the terms are getting closer to some specific value.

Now, let's suppose that the series x_n converges, i.e., it has a limit L. Then, by substituting L for both x_n and x_(n+1) in the recursive formula, we can derive an equation to solve for L.

L = 3/(4 - L)

To solve this equation, we can multiply both sides by (4 - L):

L(4 - L) = 3

Expanding and rearranging:

4L - L^2 = 3

Rearranging again:

L^2 - 4L + 3 = 0

Factoring:

(L - 1)(L - 3) = 0

This gives us two potential solutions: L = 1 and L = 3. However, we still need to determine which one is the actual limit of the series x_n.

To do that, let's analyze the behavior of the terms in the sequence as n approaches infinity.

As n gets larger, we can observe that the terms in the sequence alternate between values greater than 1 and values less than 1. This suggests that the limit of the sequence cannot be L = 1.

Therefore, the limit of the series x_n is L = 3.

Hence, we have proven that the series x_n converges, and its limit is 3.