I need some help with a calculus problem?

Hi

Since a_1 = 2, If we can show that a_(n + 1) ≤ a_n, we'll know that a_n ≤ 2. Let y = a_(n + 1) and x = a_n.

a_(n + 1) = 1/(3 - a_n)

y = 1/(3 - x)

y/x = 1/(3/x - 1)

y/x is negative if x > 3, or a_n > 3.

We can use induction to show that 0 < a_n ≤ 2.

a_1 ≤ 2, so we must show that:

a_(k + 1) ≤ 2 is also true. Since a_(k + 1) = 1/(3 - a_k), we can also write this as:

1/(3 - a_k) ≤ 2

1/(3 - a_k) ≤ 2

3 - a_k ≥ 1/2

-a_k ≥ -5/2

a_k ≤ 5/2 ≤ 2

To show that a_n > 0:

a_1 > 0

Then we must show that

a_(k + 1) > 0

1/(3 - a_k) > 0

3 - a_k < ∞

Since that last inequality is obviously true, a_(k + 1) > 0 must be true. We have showed that 0 < a_n ≤ 2.

Now we must show that a_n is decreasing. To do this, we have to show that, a_(k + 1)/a_k < 1.

a_(k + 1)/a_k

= [1/(3 - a_k)]/a_k

= 1/(3/a_k - 1)

= 1/(3/b - 1) <--- substitute b = a_k

Since we've shown that 0 < b < 2, from which it is obvious that 0 < b < 3, 3/b must be greater than 1. From that, it is obvious that 1/(3/b - 1) < 1. So the sequence is decreasing.

To find the limit of the sequence, take the limit as n increases without bound of the recursive equation.

a_(n + 1) = 1/(3 - a_n)

lim (n->∞) [a_(n + 1)] = lim (n->∞) [1/(3 - a_n)]

Since 0 < a_n ≤ 2 and a_n is decreasing, a_(n + 1) approaches a_n as n increases without bound. Let a_n approach L.

lim (n->∞) [a_(n + 1)] = lim (n->∞) [1/(3 - a_n)]

L = 1/(3 - L)

L(3 - L) = 1

3L - L^2 = 1

L^2 - 3L + 1 = 0

L = [3 ± √(9 - 4)]/2

= [3 ± √5]/2

Since [3 + √5]/2 > 2, the solution must be L = [3 - √5]/2 ≈ 0.382.

I hope this helps!