Answer
Limit: $\frac{3+\sqrt 5}{2}$
Work Step by Step
1) Let's show that \( a_n \) is increasing:
Let's show that \( a_n > 0 \) for all \( n \).
For \( n = 1 \), \( a_1 > 0 \).
Suppose that \( a_n > 0 \), and show that \( a_{n+1} > 0 \).
We have \( a_n > 1 \), then \( \frac{1}{a_n} < 1 \), then \( -\frac{1}{a_n} > -1 \), then \( 3 - \frac{1}{a_n} > 2 > 0 \).
So \( a_n > 0 \) for all \( n \).
Also, we have \( a_1 < a_2 \) because \( a_1 = 1 \) and \( a_2 = 2 \).
Suppose that \( a_n < a_{n+1} \):
And let's show that \( a_{n+1} < a_{n+2} \):
We have \( a_n < a_{n+1} \), so \( \frac{1}{a_n} + 1 < \frac{1}{a_n} \) (because \( a_n > 0 \)).
Then \( -\frac{1}{a_n} < -\frac{1}{a_{n+1}} \).
Then \( 3 - \frac{1}{a_n} < 3 - \frac{1}{a_{n+1}} \).
So \( a_{n+1} < a_{n+2} \).
Therefore, \( a_n \) is increasing.
2) Let's show that \( a_n < 3 \):
For \( n = 1 \), \( a_1 < 3 \).
Suppose that \( a_n < 3 \) and show that \( a_{n+1} < 3 \):
Since \( a_n > 0 \), therefore \( -\frac{1}{a_n} < 0 \).
So \( 3 - \frac{1}{a_n} < 3 \).
So \( a_{n+1} < 3 \).
Thus, for every \( n \), \( a_n < 3 \).
3) If \( a_n \) is increasing and increased by 3, then \( a_n \) converges.
Let's denote \( \lim_{n \to \infty} a_n = L \).
So \( L = 3 - \frac{1}{L} \).
So L=(3+$\sqrt 5$ )/2
