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.