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Math 341 Lecture
§2.4: The Monotone Convergence Theorem
and a First Look at Infinite Series
Not all bounded sequences, like (−1)n, converge, but if we knew the bounded sequence was monotone, then this would change.
Definition 2.4.1. A sequence (an) is increasing if an ≤ an+1 for all n ∈ N and decreasing
if an ≥ an+1 for all n ∈ N. A sequence is monotone if it is either increasing or decreasing.
Theorem 2.4.2 (Monotone Convergence Theorem). If a sequence is monotone
and bounded, then it converges.
Proof. Suppose (an) is monotone and bounded.
To show convergence by the -N definition, we will need to “guess” the limit s.
We will assume the sequence is increasing (the decreasing case is similar).
The sequence (an) gives a set of points {an : n ∈ N} which by hypothesis is bounded.
As (an) is increasing, we take as our “guess” for the limit,
s = sup{an : n ∈ N}.
Now we let > 0 and seek for N.
Because s is the least upper bound of {an : n ∈ N}, the number s − is not an upper bound, and so there is a number aN in the sequence for which
s − < aN.
That is, we have found an N that goes with .
Since (an) is increasing, we know that aN ≤ an for all n ≥ N.
With s being an upper bound we can obtain
s − < aN ≤ an ≤ s < s + for all n ≥ N.
This of course is nothing more than |an − s| < for all n ≥ N.
The Monotone Convergence Theorem asserts the convergence of a sequence without knowing what the limit is!
There are some instances, depending on how the monotone sequence is defined, that we can get the limit after we use the Monotone Convergence Theorem.
Example. Recall the sequence (xn) defined inductively by
x 1 = 1, xn+1 = (1/2)xn + 1, n ∈ N.
One uses induction to show that (xn) is increasing: xn ≤ xn+1 for all n ∈ N.
One also uses induction to show that xn ≤ 2 for all n ∈ N.
Thus (xn) is monotone and bounded.
By the Monotone Convergence Theorem, (xn) converges, to a number, say s.
Can we find what it converges to?
Well if s = lim xn then does lim xn+1 exist? what is it?
The “new” sequence (xn+1) is almost the “old” sequence (xn) except it is missing the first term of (xn).
So (xn+1) converges to the same thing as (xn):
nlim→∞ xn+1^ =^ s. Now by the Algebraic Limit Theorem, we have
s = lim n→∞ xn+1 = (1/2) lim n→∞ xn + 1 = (1/2)s + 1,
where we thought of the constant 1 as the constant sequence of 1’s.
We solve s = (1/2)s + 1 for s to get the limit of the sequence:
s = 2.
The Monotone Convergence Theorem is extremely useful in the study of infinite series.
Definition 2.4.3. For a sequence (bn), an infinite series is
∑^ ∞
n=
bn = b 1 + b 2 + b 3 + · · ·.
The sequence of partial sums of an infinite series is
sm =
∑^ m
n=
bn = b 1 + b 2 + · · · + bm.
The infinite series
n=1 bn^ is said to^ converge^ to^ B^ if the sequence of partial sums (sm) converges to B, and we write
B =
∑^ ∞
n=
bn.
Otherwise, if (sm) diverges, then the infinite series diverges as well.
Example 2.4.4. The partial sums for the infinite series
∑^ ∞
n=
n^2
are
sm = 1 +
m^2 < 1 +
m(m − 1) = 1 +
m − 1
m
= 1 + 1 − (1/m) < 2.
Appendix. Proof of Theorem 2.4.6.
Suppose that
n=1 2 nb 2 n converges.
So its sequence of partial sums
tk = b 1 + 2b 2 + 4b 4 + · · · 2 kb 2 k
converges, and hence is bounded; there is M > 0 such that tk ≤ M for all k ∈ N.
Now because bn ≥ 0, the sequence of partial sums (sm) for
n=1 bn, is increasing. To show that
n=1 bn^ converges we need only show that (sm) is bounded. Fix m and let k be large enough to ensure that
m ≤ 2 k+1^ − 1.
Then sm ≤ s 2 k+1− 1
and
s 2 k+1− 1 = b 1 + (b 2 + b 3 ) + (b 4 + b 5 + b 6 + b 7 ) + · · · + (b 2 k + · · · + b 2 k+1− 1 ) ≤ b 1 + (b 2 + b 2 ) + (b 4 + b 4 + b 4 + b 4 ) + · · · + (b 2 k + · · ·b 2 k ) = b 1 + 2b 2 + 4b 4 + · · · + 2kbk = tk.
Thus sm ≤ tk ≤ M , and so (sm) is bounded.
By the Monotone Convergence Theorem, (sm) converges, and then so does
n=1 bn. The opposite direction is shown by its contrapositive: if
n=1 2 nb 2 n (^) diverges then ∑∞ n=1 bn diverges.
This argument is very much like that used for the harmonic series in Example 2.4.5.
