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Monotone Convergence Theorem
Motivation: As we mentioned previously, sometimes it is difficult to “guess” the limit of a sequence and prove that it is convergent. So, we hope that we can find some other clues to determine if a sequence converges or not. Monotone Convergence Theorem is one of the methods to check the convergence of a sequence or series.
Definition ( Upper Bound and Lower Bound ) For a sequence where cn R , if cn M for all n , then M is known as an upper bound of the sequence . On the other hand, if there exists L such that L cn for all n, then L is known as a lower bound of the sequence. Among all the upper bounds, the smallest one is known as least upper bound ( LUB ). Similarly, among all the lower bound, the largest one is known as greatest lower bound.
A bounded sequence means there exist K such that | cn | K for all n.
Theorem ( Monotone Convergence Theorem ) For a sequence and cn R , if cn cn+1 and cn < M for all n N, then converges.
Proof: cn < M for all n M. Thus, there exists L such that L is the least upper bound. In other words, for any > 0, the following statement holds
1. Between L and L+ , none of the will fall into this interval because L is an upper bound. 2. Between L- and L, there must exist at least a ck in this interval. Otherwise, L- is an upper bound of such that L- < L. But we know that L is the least upper bound. So, that condition will never happen.
Thus, we have L- < ck < L+ . Furthermore, ck cn for n k, and cn < L+ for all n because (L+ ) is an upper bound L- < cn < L+ for n k .
In other words, for every > 0 , there exists k such that | cn – L | < for n k
That is how we define the limit of a sequence. So, converges.
Similarly, for a decreasing sequence, if it has a lower bound, then it converges.
Example: Sk = , k N . Prove that converges. Proof: Sk = =
Thus, Sk > 0 for every k N. Furthermore, Sk > Sk+1 for every k N.
The sequence has lower bound 0 and it is decreasing. From monotone convergence theorem, this sequence converges.
Theorem: Every convergent sequence is bounded.
Proof: If , that means For every > 0, there exists a corresponding K such that | cn – c | < for n > K - < cn – c < c - < cn < c + for n > K
Let M = max { | c - |, | c+ | } . | cn | < M for n > K
Let B= max{ | c1 |, | c2 |, | c3 | , …, |cK | } . So, | cn | < B for n K .
Let U = max { M, B }. We have | cn | < U for every n . In other words, it is bounded.
Theorem: If and , then (1) (2) (3) (4) if c 0 .
Proof: and . That means
for every > 0 , there exist K1 such that | bn – b | < for n < K1
and for every > 0, there exist K2 such that | cn – c | < for n < K2
Let K=max( K1, K2 ) .
(1) | bn – b | + | cn – c| | (bn –b) + (cn – c) | = | (bn+cn) – (b+c) | So, for n > K, | (bn+cn) – (b+c) | | bn – b | + | cn – c| < +
In other words, for every >0, we can properly choose and such that = + , and there exists a corresponding K to let the following statement be true
| (bn+cn) – (b+c) | < , for n > K
Thus,
(2) | bn – b | + | cn – c| | (bn – b) + ( c – cn ) | = | ( bn – cn) – (b-c) | So, for n > K , | ( bn – cn) – (b-c) | | bn – b | + | cn – c| < +
In other words, for every >0 , we can properly choose and such that = + , and there exists a corresponding K such that | ( bn – cn) – (b-c) | < for n > K .
Thus,
(3) is a convergent sequence. So, there exists M such that | cn | < M And | bn cn – bc | = | bncn – bcn + bcn – bc | | cn | | bn – b | + | b| | cn – c| M | bn – b | + |b| |cn – c|
So, for n>K, we have | bn cn – bc | M | bn – b | + |b| |cn – c | M + |b|
Hence, for every > 0 , we can properly choose and such that = M + |b| , and there exists a corresponding K such that
| bn cn – bc | < for n < K .
In other words, .
(4) and c 0 . Thus, there only exist finitely many terms from {cn } such that cn = 0. So, we can exclude those finitely many terms from {cn} to define a new sequence { dn } such that
dn = Then . ( It can be easily proved ).
Use the result from (3),
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