Sometimes you cannot compute a sequence's terms in closed form, as with $a_{n+1} = \sqrt{2 + a_n}$. The monotone convergence theorem proves such a sequence converges without knowing the limit in advance, using completeness; after that, the limit laws find the limit. You met bounded monotone sequences in Calculus 2 as a fact; here is the proof.
Key idea
Monotone convergence theorem. A bounded increasing sequence converges to $\sup\{a_n\}$; a bounded decreasing sequence converges to $\inf\{a_n\}$. For a recursion $a_{n+1} = g(a_n)$ with $g$ continuous: prove the sequence is monotone and bounded (usually by induction), and only then solve $L = g(L)$.
Why it works
Let $(a_n)$ be increasing and bounded above, and let $s = \sup\{a_n\}$, which exists by completeness. Given $\varepsilon > 0$, the number $s - \varepsilon$ is not an upper bound, so some $a_N > s - \varepsilon$. Because the sequence increases, every $n \ge N$ has $s - \varepsilon < a_N \le a_n \le s$, so $|a_n - s| < \varepsilon$. $\blacksquare$ The theorem says nothing about the value of $s$; the recursion supplies it, because $a_{n+1} \to s$ as well, so $s = g(s)$.
Worked examples
Worked example: A nested square root
Let $a_1 = 0$ and $a_{n+1} = \sqrt{2 + a_n}$. Show that $(a_n)$ converges and find its limit.
- Bounded: $a_1 < 2$, and if $a_n < 2$, then $a_{n+1} = \sqrt{2 + a_n} < \sqrt{4} = 2$. By induction, $a_n < 2$ for all $n$.
- Increasing: for $0 \le x < 2$, $x^{2} < x + 2$ (since $x^{2} - x - 2 = (x - 2)(x + 1) < 0$), so $x < \sqrt{2 + x}$. Hence $a_n < a_{n+1}$.
- So the sequence converges, to some $L$ with $0 \le L \le 2$. Then $L = \sqrt{2 + L}$, so $L^{2} - L - 2 = 0$ and $L = 2$ or $L = -1$. Since $L \ge 0$, $L = 2$.
Answer: $a_n \to 2$.
Worked example: A linear recursion
Let $a_1 = 0$ and $a_{n+1} = \frac{1}{2}a_n + 3$. Find $a_2$, $a_3$ and the limit.
- $a_2 = 3$, $a_3 = \frac{9}{2}$. The terms increase and stay below $6$ (if $a_n < 6$ then $a_{n+1} < 3 + 3 = 6$).
- The limit satisfies $L = \frac{1}{2}L + 3$, so $L = 6$.
Answer: $a_2 = 3$, $a_3 = \frac{9}{2}$, limit $6$.
Common mistake
Solving $L = g(L)$ first. For $a_1 = 1$ and $a_{n+1} = 2a_n + 1$, the equation $L = 2L + 1$ gives $L = -1$, but the terms are $1, 3, 7, 15, \ldots$: there is no limit. The equation only says what a limit would have to be. Also, when the equation has two roots, use what you proved about the terms (all $\ge 0$, all below the smaller root, and so on) to choose the right one.