## Monotone Convergence

We start with a useful theorem.

Monotone Convergence Theorem (MCT). A sequence ismonotonically increasing(or justincreasing) if for all n. Now the theorem says: an increasing sequence with an upper bound is convergent.

**Proof**.

Let *L* = sup{*a*_{1}, *a*_{2}, … }, which exists by completeness of **R**. Thus, each *a _{n}* ≤

*L*. On the other hand, for each ε>0, since

*L*-ε is not an upper bound of the {

*a*}, we can find

_{n}*a*>

_{N}*L*-ε. Hence for all

*n*>

*N*, we have

*a*≥

_{n}*a*>

_{N}*L*-ε while

*a*≤

_{n}*L*which gives:

♦

In practice, the monotonic convergence theorem often tells us a limit exists without giving any hint on what it might be. In some cases, it may be derivable.

**Example 1**. Consider the recurrence relations , . Prove that (*x _{n}*) is convergent and find its limit.

**Answer**. We shall prove that by induction. For *n*=0 this is obvious. Suppose we have . For the next iteration, we have:

This shows that which completes the induction. Next, note that since *x _{n}* > 0. Hence, (

*x*) is an increasing sequence with an upper bound and must converge to some

_{n}*L*.

From , the LHS → *L* while the RHS → 6-(5/*L*). Equating and solving give *L*=1 or *L*=5. Since each *x _{n}* ≥ 2, the limit can’t be 1. So

*L*=5. ♦

**Example 2**. Prove that converges.

**Answer**. Let *a _{n}* be the

*n*-th partial sum. Since

*n*! ≥ 2

^{n-1}, we have . But converges by sum of geometric series. Hence,

*a*is an increasing sequence with an upper bound. By MCT, it converges. ♦

_{n}However, there’s no easy way to evaluate this sum.

*Note: in proving MCT, we explicitly used the fact that R is complete. This property is critical since the theorem does not hold for Q. For example, we can easily have the sequence 1, 1.4, 1.41, 1.414, … which contains more and more significant figures of √2. We get an increasing sequence of rational numbers, with no limit in Q. On the other hand, since R is complete, there’s a limit in R.*

This gives a good sense of what completeness means. We get a mental picture of “gaps” occurring in **Q**, so that a sequence of elements which get successively closer still may not converge. The next section will elucidate this even further.

## Cauchy Sequences

We’ll define what it means for a sequence of numbers to get successively closer.

Definition. ACauchy sequenceis a sequence (a_{n}) where

- for any ε>0, there exists N such that when m, n > N, we have |a
_{m }– a_{n}| < ε.

Some results are quite easy to prove.

- A convergent sequence is Cauchy.
- A Cauchy sequence is bounded.

**Proof**.

For the first statement, suppose the sequence → *L*. For any ε>0, there exists *N* such that when *n* > *N*, we have |*a _{n}* –

*L*| < ε/2. Hence, when

*m*,

*n*>

*N*, we have . So the sequence is Cauchy.

For the second, pick ε=1. We get an *N* such that when *m*, *n* > *N*, we get |*a _{m}* –

*a*| < 1. Fix some

_{n}*n*>

*N*and let

*B*= |

*a*|. It follows that for all

_{n}*m*>

*N*, |

*a*| ≤ |

_{m}*a*–

_{m}*a*| + |

_{n}*a*| <

_{n}*B*+1. So the set of

*a*for

_{m}*m*>

*N*is bounded. Since there’re only finitely many terms up to

*N*, the whole sequence is bounded. ♦

Notice that the property of being a Cauchy sequence is inherent in the sequence and does not rely on the embedding. In short, a sequence of rational numbers is Cauchy regardless whether we consider it embedded in **Q** or in **R**. On the other hand, whether a sequence of rational numbers converges depends on whether the ambient space is **Q** or **R**.

Theorem. A Cauchy sequence of real numbers converges to a real number.

**Proof**. Suppose (*a _{n}*) is Cauchy. In particular, it’s bounded and we can let . Then since

*b*takes the sup of more elements than

_{n}*b*. Since (

_{n+1}*a*) has a lower bound so does (

_{n}*b*) and by MCT, (

_{n}*b*) converges to some

_{n}*L*.

It remains to show (*a _{n}*) →

*L*. Let ε > 0. Find integer

*N*such that:

- when
*m*,*n*>*N*, we have |*a*–_{m}*a*| < ε/2;_{n} - when
*n*>*N*, we have |*b*–_{n}*L*| < ε/2, or*L*-(ε/2)*<**b*<_{n }*L*+(ε/2).

Now since *L*-(ε/2)* <* *b _{N}*

_{+1}, and sup is the minimum upper bound,

*L*-(ε/2) is not an upper bound of so we can find

*n*>

*N*such that . Yet , which gives .

Hence, for all *m* > *N*, we have . ♦

## Squeeze Theorem

The following theorem is simple but handy.

Squeeze Theorem. Let be three sequences with . Suppose a_{n}→ L and c_{n}→ L. Then b_{n}→ L.

**Proof**. Let ε > 0. Pick *N* such that whenever *n* > *N* :

- ;
- ;

Hence when *n* > *N*, we have and , i.e. . ♦

**Example 3**. Prove that has a limit of 0.

**Proof**. Since sin(*n*) ≤ 1, we have 2^{sin(n)} ≤ 2 and . Since 2/*n*^{2} → 0, the result follows from the squeeze theorem. ♦

## Cesàro mean

**Exercise (hard)**. Prove that if *a _{n}* →

*L*, and , then

*b*→

_{n}*L*.

**Sketch of answer** (highlight to read). [ For ε > 0, pick *N* such that if *n*>*N*, |*a _{n}*–

*L*| < ε/2. Then |

*b*–

_{n}*L*| ≤ (|

*a*–

_{1}*L*|+|

*a*–

_{2}*L*|+…+|

*a*–

_{n}*L*|)/

*n*. Let

*B=|*|. Then |

*a*–_{1}*L*|+|*a*–_{2}*L*|+…+|*a*-L_{N}*b*–

_{n}*L*|≤(

*B*/

*n*) + ε/2. So pick

*M*>max(

*N*, 2

*B*/ε). Then when

*n*>

*M*,

*B*/

*n*< ε/2 also. ]

The new sequence which is obtained via taking the mean of the first *n* terms is called the Cesàro mean. The above exercise tells us taking the Cesàro mean preserves the limit if it already exists. On the other hand, some non-converging sequences converge after taking the Cesàro mean. E.g. if *a _{n}* = (-1)

^{n}, then its Cesàro mean converges to 0 since the

*n*-th term is bounded by [-1/

*n*, +1/

*n*], so it converges by the squeeze theorem.

## Subsequences

A **subsequence** of a sequence (*a _{n}*) is a sequence of the form (

*b*), where

_{m}, for some increasing *n*(1) < *n*(2) < *n*(3) < …

Thus in the sequence *a _{n}* = (-1)

^{n}, the even terms form a subsequence (+1, +1, +1, …) with indices

*n*(1)=2,

*n*(2)=4,

*n*(3)=6, … . The odd terms form a subsequence (-1, -1, -1, …) with indices

*n*(

*k*)=2

*k*-1. Hence, a non-convergent sequence can have subsequences which converge to different values.

**Theorem**. Let (*a _{n}*) be a sequence.

- If lim (
*a*) =_{n}*L*, then every subsequence converges to*L*. - If every subsequence converges to the same value
*L*, then lim (*a*) =_{n}*L*.

**Proof**.

For the first statement, suppose we have a subsequence (*b _{m}*), where ,

*n*(1) <

*n*(2) <

*n*(3) < … . Let ε>0. There exists

*N*such that when

*n*>

*N*, |

*a*–

_{n}*L*| < ε. Since

*n*(

*m*) ≥

*m*, whenever

*m*>

*N*, we also have . Hence the subsequence (

*b*) also converges to

_{m}*L*.

For the second statement, suppose (*a _{n}*)

*doesn’t converge to L*. Unwinding the definition, there exists ε>0 such that for each

*N*, there exists

*n*>

*N*such that |

*a*–

_{n}*L*

*| ≥*ε. In short, there are infinitely many

*n*for which |

*a*–

_{n}*L*

*| ≥*ε. If we pick this subsequence, it cannot possibly converge to

*L*. ♦