Example 1: (a) Determine whether the sequence defined as follows is convergent or divergent. a_1 = 1, a_{n+1} = 4 - a_n for n\geq 1. (b) What happens if the first term is a_1 = 2?
Hint: (a) It a sequence of alternating 1 and 3. (b) It is a constant sequence.
Example 2: Suppose you know that \{a_n\} is a decreasing sequence and all its terms lie between 5 and %8. Explain why the sequence has a limit. What can you say about the value of the limit?
Hint: Use the monotonic sequence theorem and the sandwich theorem.
Example 3: Determine whether the sequence is increasing, decreasing or not monotonic. Is the sequence bounded? (a) a_n = (-2)^{n+1}; (b) a_n = 1/(2n+3); (c) a_n = (2n-3)/(3n+4); (d) a_n = n + 1/n.
Hint: (b) 0 < a_n < 1; (c) -1 < a_n < 1 and a_n = \frac{2}{3} - \frac{17}{3(3n+4)}; (d) a_{n+1} - a_n = 1 - \frac{1}{n(n+1)} > 0.
Example 4: Show that the sequence defined by a_1 = 1, a_{n+1} = 3 - 1/a_n is increasing and a_n < 3 for all n. Deduce that \{a_n\} is convergent and find its limit.
Solution: First of all, we want to show that 1 \leq a_n < 3 for all n. Since a_1 already satisfies the hypothesis, it suffices to show if 1 \leq a_n < 3, so is a_{n+1} (see the remark below for further discussion). Since a_{n+1} = 3 - 1/a_n and 1 \leq a_n < 3, 1 < 3 - 1/1 \leq 3 - 1/a_n < 3 - 1/3 < 3, hence 1 \leq a_{n+1} < 3. Secondly, since a_1 = 1 < 2 = a_2, to show that a_n < a_{n+1}, it suffices to show that if a_{n-1} < a_n then a_n < a_{n+1} for all n > 1 (again see the remark below). Because 0 < a_{n-1} < a_n, we know that a_n = 3 - 1/a_{n-1} < 3 - 1/a_n = a_{n+1}. This ends the proof of the monotonicity. Because the sequence is monotonic and bounded, it converges by the monotonic sequence theorem. Let L be its limit. Then 1\leq L\leq 3 and L = 3 - 1/L. Solve the equation and get L = \frac{3+\sqrt{5}}{2}.
Remark: The idea used in the proof is mathematical induction which is commonly used in mathematical proof.