Ex. (n)_n is increasing. (1/n)_n is decreasing.

Monotone Convergence Theorem. If a sequence is both bounded and monotone, it converges.
Proof. Assuming a_n is an increasing bounded sequence, then since a_n is bounded, {a_n : n } has a least upper bound s = sup{a_n : n }. Let > 0 be given. By the definition of the supremum, there is an a_n where s - < a_n < s. Since a_n is increasing, there exists an N such that if n > N, then s - < a_N < a_n < s. This inequality states that |a_n - s| < . Therefore, a_n converges (a_n s, in this case actually). A similar proof can be designed for a decreasing sequence with the infimum.//

The Monotone Convergence Theorem states that a sequence will converge without any notion of a limit.

Uber Long Example. Prove that the sequence (a_n)_n converges and find its limit, where a_n is defined as:

a₁ = 1
a_n+1 = (20 + a_n) / 5

To prove that this thing converges, we only have to show that it's monotone and bounded. First let's show that it's increasing. Ie, we want to prove that a_n < a_n+1 for all n after some point.

a₁ = 1
a₂ = (20 + a₁) / 5 = 21/5 = 4.2

For a basis case, a₁ < a₂.

So now we can say that for at least one n, a_n < a_n+1. If we can prove that the next case applies, then it will work for all n afterwards. (picture Prof. Grimaldi's famous example, the falling dominos).

By the definition of the sequence, a_n+1 = (20 + a_n) / 5. Rearranging this to solve for a_n yields, a_n = 5*a_n+1 - 20. Substituting this into a_n < a_n+1 gives us:

5*a_n+1 - 20 < a_n+1.

If we manipulate this a bit, add 20 to both sides and then divide by five, notice that we can get...

a_n+1 < (20 + a_n+1) / 5.

And by the definition of the sequence, the right side is actually the next number in the sequence.

a_n+1 < a_n+2

So, without cheating or doing any crap, simply by rewriting an in an alternate form allowed for the next number in the sequence to follow through. Since the basis case was n = 1, and it works for all n > 1, we can safely declare this sequence to be increasing.

Now, for the Monotone Convergence Theorem to work, we need to prove that it is both increasing and bounded. We know that it starts at 1, and it gets bigger for every number in the sequence after that, but do we know if there is a max value?

Since is was already given that a_n < a_n+1, we can use the definition of the sequence to rewrite it as a_n < (20 + a_n) / 5. Multiplying by 5 to both sides gives 5*a_n < 20 + a_n. Subtracting by a_n and then dividing by 4 gives a very nice, simple response to our question as to whether a_n is bounded or not: a_n < 5. And, since our premise that a_n < a_n+1, our claim that a_n < 5 is likewise good for any n.

Since a_n begins at 1 and is increasing and is always less than 5, we can safely declare that the sequence a_n is bounded by M = 5. Since a_n is both bounded and monotone, we can invoke the Monotone Convergence Theorem and declare that a_n converges to some limit L.

Since we know that a_n = L,

by the definition of the limit we can also say that a_n+1 = L.

Using this convenient relationship, a_n = a_n+1.

This implies that L =  (20 + L) / 5. Multiplying by 5, subtracting by L, and dividing by four (notice very similar to the bounded proof), yields L = 5. Surprise surprise, the bounded maximum was also limit to the sequence. While this doesn't happen all the time, it's pretty typical.

If a sequence is increasing, always positive, and bounded by M, it'll probably converge to M. Sequences can do weird things though, like eventually increase or be bounded but converge to e, or something.