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*| is greater or equal to ε. Suppose it's monotone, without loss of generality

_{n}*nondecreasing*; that implies that for all

*N*, we can find

*n*>

*m*≥

*N*such that

*a*–

_{n}*a*≥ ε. Now suppose our sequence is bounded above by some bound

_{m}*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

*a*. We can find points later in the sequence that are separated from each other by at least ε, but if we do this ⌈(

_{1}*B*–

*a*)/ε⌉ times, then we'll have found a sequence point greater than the alleged bound.

_{1}
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