If a sequence of real numbers (a_n) is bounded and monotone (and I'm actually going to say nondecreasing, without loss of generality), then it converges. I'm going to tell you why and I'm going to tell you twice.

If our sequence is bounded, the completeness of the reals ensures that it has a least upper bound, which we'll call, I don't know, B, but there have to be sequence elements arbitrarily close to (but not greater than) B, because if there weren't, then B couldn't be a least upper bound. So for whatever arbitrarily small ε, there's an N such that a_N > B – ε, which implies that |a_N – B| < ε, but if the sequence is nondecreasing, we also have |a_n – B| < ε for n ≥ N, which is what I've been trying to tell you—

—twice; suppose by way of contraposition that our sequence is not convergent. Then there exists an ε such that for all N, there exist m and n greater or equal to N, such that |a_m – a_n| is greater or equal to ε. Suppose it's monotone, without loss of generality nondecreasing; that implies that for all N, we can find n > m ≥ N such that a_n – a_m ≥ ε. Now suppose our sequence is bounded above by some bound B. However, we can actually describe an algorithm to find sequence points greater than B, thus showing that this alleged bound is really not a bound at all. Start at a₁. We can find points later in the sequence that are separated from each other by at least ε, but if we do this ⌈(B – a₁)/ε⌉ times, then we'll have found a sequence point greater than the alleged bound.