Using the Monotone Convergence Theorem to Prove a Recursive Sequence Converges: Example 1

1. Formula
   Observation

   The board shows the sequence term written as $a_n$ in the recurrence and in later computations.

2. Audio
   Observation

   The lecturer refers to the recursively defined sequence and its terms while writing $a_n$.

$a_n$

The n-th term of the recursively defined real sequence under study.

$n \ge 1$

1. Formula
   Observation

   The top line on the board reads $a_{n+1}=\frac{1}{3}(a_n+1)$.

2. Audio
   Observation

   The lecturer says, "Consider a sub n plus 1 is equal to $1/3$ of a sub n plus 1."

$a_{n+1}$

The next term of the sequence, defined from the previous term by the given recurrence.

$n \ge 1$

1. Formula
   Observation

   The board includes the condition $n \ge 1$ beside the recurrence.

2. Audio
   Observation

   The lecturer says the recurrence is true for "n bigger than or equal to 1".

n

Index variable for the sequence terms.

$n \ge 1$

1. Formula
   Observation

   The board shows the initial condition $a_1=1$.

2. Audio
   Observation

   The lecturer says, "with a sub 1 being equal to ... a sub 1 is equal to 1".

$a_1$

The first term of the sequence, used as the base case for computations and induction.

fixed initial value

1. Formula
   Observation

   The board shows $a_2=\frac{1}{3}[1+1]=\frac{2}{3}$.

2. Audio
   Observation

   The lecturer computes $a_2$ verbally as one-third of one plus one, giving two-thirds.

$a_2$

The second term of the sequence, obtained by substituting $a_1=1$ into the recurrence.

specific sequence term

1. Formula
   Observation

   The board shows $a_3=\frac{1}{3}\left[\frac{2}{3}+1\right]=\frac{5}{9}$.

2. Audio
   Observation

   The lecturer computes $a_3$ verbally and obtains five-ninths.

$a_3$

The third term of the sequence, obtained by substituting $a_2=\frac{2}{3}$ into the recurrence.

specific sequence term

1. Formula
   Observation

   The board writes "Claim: $a_n \ge \frac{1}{2}$".

2. Audio
   Observation

   The lecturer states that the first claim is that the $a_n$ terms are bigger than or equal to one-half.

$a_n \ge \frac{1}{2}$

Proposed lower bound for every term of the sequence.

for all $n \ge 1$

1. Formula
   Observation

   The board writes "Suppose $a_n \ge \frac{1}{2}$" as the induction hypothesis.

2. Audio
   Observation

   The lecturer says, "suppose that a sub n is bigger than or equal to $1/2$".

$a_n \ge \frac{1}{2}$

Assumed truth of the lower-bound statement at step n in the induction proof.

inductive assumption for a fixed n

1. Formula
   Observation

   The board writes "Our sequence is bounded from below."

2. Audio
   Observation

   The lecturer concludes, "So our sequence is bounded from below."

bounded from below

Conclusion that there exists a real number L such that every sequence term satisfies $a_n \ge L$; here the exhibited lower bound is $\frac{1}{2}$.

property of the whole sequence

1. Formula
   Observation

   The board repeatedly uses $a_n$, $a_{n+1}$, and $a_1=1$ in the recurrence and proof.

2. Audio
   Observation

   The lecturer refers to “our sequence $a_n$” while deriving $a_{n+1}-a_n$.

$a_n$

The recursively defined sequence under study.

Defined for $n\ge 1$ with $a_1=1$ and $a_{n+1}=\frac13(a_n+1)$.

1. Formula
   Observation

   The top line states $a_{n+1}=\frac13(a_n+1)$, and the lecturer writes $a_{n+1}-a_n$ below it.

$a_{n+1}$

The next term of the recursive sequence after $a_n$.

For $n\ge 1$.

1. Formula
   Observation

   The board shows $a_1=1$ at the top right of the initial setup.

$a_1$

Initial value of the sequence.

Fixed constant equal to $1$.

1. Formula
   Observation

   Top board line: "Consider $a_{n+1}=\frac{1}{3}(a_n+1),\ n\ge 1,\ a_1=1$".

2. Audio
   Observation

   The lecturer introduces a recursively defined sequence and reads the recurrence and initial condition aloud.

The video defines a sequence by giving a rule for the next term in terms of the current term, together with an index restriction and an initial value. The displayed rule is $a_{n+1}=\frac{1}{3}(a_n+1)$ for $n\ge 1$, with $a_1=1$. This is the object whose convergence will later be studied through the Monotone Convergence Theorem.

1. The recurrence applies for $n\ge 1$.

2. The initial value is fixed at $a_1=1$.

1. Audio
   Observation

   The lecturer says they want to use the monotone convergence theorem and therefore need to show the sequence is bounded and non-decreasing or non-increasing.

2. Diagram
   Observation

   No new formula is added yet; this is a spoken proof plan before the computations.

The lecturer explicitly frames the problem as an application of the Monotone Convergence Theorem. According to the spoken plan, to prove convergence it is enough to establish two properties of the sequence: boundedness and monotonicity. In this example the lecturer first targets boundedness from below, then announces that the next task is to show the sequence is non-increasing.

1. The method is applied to a real sequence.

2. The video states the needed ingredients as boundedness plus monotonicity, but does not write the full theorem statement on the board.

1. Recursive definition of the example sequence

1. Formula
   Observation

   Board shows $a_2=\frac{1}{3}[1+1]=\frac{2}{3}$ and $a_3=\frac{1}{3}\left[\frac{2}{3}+1\right]=\frac{5}{9}$.

2. Audio
   Observation

   The lecturer says these values give a sense of what is happening and remarks that the sequence looks like it is going down.

Before proving anything general, the lecturer substitutes the initial value into the recurrence to compute $a_2$ and $a_3$. The arithmetic shows $a_2=\frac{2}{3}$ and $a_3=\frac{5}{9}$. These concrete terms support the visual impression that the sequence is decreasing, motivating the later monotonicity check.

1. Use the recurrence $a_{n+1}=\frac{1}{3}(a_n+1)$.

2. Start from $a_1=1$.

1. Recursive definition of the example sequence

1. Formula
   Observation

   Board writes "Claim: $a_n \ge \frac{1}{2}$".

2. Audio
   Observation

   The lecturer states the first claim is that the $a_n$ terms are bigger than or equal to one-half.

The lecturer isolates a precise auxiliary statement to be proved: every term of the sequence is at least $\frac{1}{2}$. This claim is the boundedness ingredient needed for the eventual Monotone Convergence Theorem argument. The board treats it as a standalone proposition to be verified by induction.

1. Intended for all $n\ge 1$.

2. This is a proposed lower bound, not yet justified until the induction proof is completed.

1. Recursive definition of the example sequence
2. Proof strategy using the Monotone Convergence Theorem

1. Formula
   Observation

   Board writes "True if $n=1$" and then "Suppose $a_n \ge \frac{1}{2}$" followed by the implication to $a_{n+1}$.

2. Audio
   Observation

   The lecturer says the claim is clearly true and will be proved by induction.

To justify the claim $a_n\ge \frac{1}{2}$ for all $n\ge 1$, the lecturer uses mathematical induction. The structure shown on the board has a base case at $n=1$ and an inductive step that assumes the inequality at index $n$ and derives it at index $n+1$ using the recurrence. This is the standard method for turning a termwise pattern into a universal statement about the whole sequence.

1. Base case must be checked at $n=1$.

2. Inductive hypothesis is $a_n\ge \frac{1}{2}$.

3. The recurrence must be used to pass from $a_n$ to $a_{n+1}$.

1. Claimed universal lower bound for the sequence
2. Recursive definition of the example sequence

1. Formula
   Observation

   Top of board: “Consider $a_{n+1}=\frac13(a_n+1),\ n\ge1$” and $a_1=1$.

2. Formula
   Observation

   Left side already lists $a_2=\frac13(1+1)=\frac23$ and $a_3=\frac13(\frac23+1)=\frac59$.

The video studies the sequence defined by $a_{n+1}=\frac13(a_n+1)$ for $n\ge1$ with initial value $a_1=1$. The first computed terms shown on the board are $a_2=\frac23$ and $a_3=\frac59$, which motivate the later claim that all terms stay at least $\frac12$.

1. $n\ge 1$

2. $a_1=1$

1. Audio
   Observation

   “So $a_{n+1}$ minus $a_n$ … we’ll have a one third $a_n$ plus one third … and then we’ll have a minus over here $a_n$.”

2. Formula
   Observation

   The lecturer writes $a_{n+1}-a_n=\frac13a_n+\frac13-a_n=-\frac23a_n+\frac13$.

To test monotonicity, the lecturer subtracts $a_n$ from both sides of the recurrence. Substituting $a_{n+1}=\frac13(a_n+1)$ gives $a_{n+1}-a_n=\frac13a_n+\frac13-a_n$, which simplifies to $-\frac23a_n+\frac13$.

1. Use the recurrence $a_{n+1}=\frac13(a_n+1)$.

2. Algebraic simplification combines the $a_n$ terms.

1. Recursive definition of the example sequence

1. Formula
   Observation

   Blue boxed claim on board: “Claim: $-\frac23a_n+\frac13\le0 \iff \frac13\le\frac23a_n \iff \frac12\le a_n$.”

2. Audio
   Observation

   “We want to show that this is less than or equal to zero … which is exactly equivalent to saying that one third is bigger than or equal to two thirds of $a_n$ … our claim is equivalent to $a_n$ being bigger than or equal to a half, which we know is true by induction.”

The inequality $-\frac23a_n+\frac13\le0$ is algebraically equivalent to $\frac13\le\frac23a_n$, and dividing by $\frac23$ yields $\frac12\le a_n$. Thus proving $a_{n+1}-a_n\le0$ reduces to using the previously established lower bound $a_n\ge\frac12$.

1. All transformations are reversible because they use addition/subtraction and division by the positive constant $\frac23$.

2. The final implication uses the already proven fact $a_n\ge\frac12$.

1. Computing $a_{n+1}-a_n$ from the recurrence

1. Formula
   Observation

   Board writes $a_{n+1}-a_n\le0\Rightarrow a_{n+1}\le a_n$ and circles $a_{n+1}\le a_n$.

2. Audio
   Observation

   “And so now I have that $a_{n+1}$ minus $a_n$ is less than or equal to zero, which says that $a_{n+1}$ is less than or equal to $a_n$, and this is the exact thing that says our sequence is non-decreasing.”

1. The spoken phrase “non-decreasing” conflicts with the displayed inequality $a_{n+1}\le a_n$ and the later written phrase “bounded and non-increasing”; the mathematical content on the board indicates non-increasing.

Once $-\frac23a_n+\frac13\le0$ is established, the lecturer concludes $a_{n+1}-a_n\le0$, hence $a_{n+1}\le a_n$. This is the criterion for a non-increasing sequence. The board later explicitly writes that $\{a_n\}_{n=1}^\infty$ is bounded and non-increasing.

1. Valid for every $n\ge1$ once the lower bound $a_n\ge\frac12$ is known.

1. Equivalence between nonpositivity of the difference and the bound $a_n\ge\frac12$

1. Formula
   Observation

   Board writes “$\{a_n\}_{n=1}^\infty$ bounded and non-increasing” and then “Monotone Convergence $\Rightarrow a_n\to L$.”

2. Audio
   Observation

   “And so this allows us to use the monotone convergence theorem … implies that the $a_n$ converge to some number $L$.”

Because the sequence has been shown to be bounded below and non-increasing, the lecturer invokes the Monotone Convergence Theorem to conclude that $a_n$ converges to some real limit $L$.

1. The sequence must be monotone and bounded.

2. Here boundedness below was already shown earlier on the board, and monotonicity is non-increasing.

1. From $a_{n+1}-a_n\le0$ to $a_{n+1}\le a_n$

1. Formula
   Observation

   Board writes $L=\frac13(L+1)\Rightarrow \frac23L=\frac13\Rightarrow L=\frac12$.

2. Audio
   Observation

   “The left hand side will converge to $L$. The right hand side will converge to one third of $L$ plus one … two thirds of $L$ is equal to one third, and therefore $L$ is equal to one half.”

After knowing $a_n\to L$, the lecturer passes to the limit in the recurrence $a_{n+1}=\frac13(a_n+1)$. Since both $a_n$ and $a_{n+1}$ tend to $L$, the limiting equation is $L=\frac13(L+1)$, which simplifies to $\frac23L=\frac13$ and hence $L=\frac12$.

1. Requires that $a_n$ converges, which was established by the Monotone Convergence Theorem.

2. Uses continuity of the affine map $x\mapsto \frac13(x+1)$ implicitly by taking limits termwise.

1. Application of the Monotone Convergence Theorem

1. Formula
   Observation

   Board states "Claim: $a_n \ge \frac{1}{2}$" and later concludes "Our sequence is bounded from below."

2. Audio
   Observation

   The lecturer says the claim is proved by induction and concludes the sequence is bounded from below.

For the recursively defined sequence with $a_{n+1}=\frac{1}{3}(a_n+1)$ and $a_1=1$, one has $a_n\ge \frac{1}{2}$ for every $n\ge 1$; hence the sequence is bounded from below.

1. $a_{n+1}=\frac{1}{3}(a_n+1)$ for $n\ge 1$.

2. $a_1=1$.

for all $n \ge 1$

1. Audio
   Observation

   The lecturer says, "Now let's show the sequence is non-increasing. So how do we do that? So let's look at the difference."

2. Diagram
   Observation

   No formula for the difference is actually written before the clip ends.

1. The clip ends before the non-increasing argument is carried out.

2. Only the intention to inspect a difference is audible; the actual expression is not shown.

The lecturer announces that the next objective is to show the sequence is non-increasing and begins by saying they will look at the difference.

1. The previously proved lower bound is already established.

2. The same recursively defined sequence is under consideration.

video states the intended next claim but does not complete it within this clip

1. Formula
   Observation

   Green text on board: “Our sequence is bounded from below.”

2. Formula
   Observation

   The induction block above states “Claim: $a_n\ge\frac12$”, “True if $n=1$”, and the inductive step using $a_{n+1}=\frac13(a_n+1)\ge\frac13(\frac12+1)=\frac12$.

For all $n\ge1$, $a_n\ge\frac12$, so the sequence is bounded from below.

1. $a_{n+1}=\frac13(a_n+1)$ for $n\ge1$

2. $a_1=1$

3. Induction has already been carried out on the board before this clip’s main derivation.

For every integer $n\ge1$.

1. Formula
   Observation

   Board writes $a_{n+1}-a_n\le0\Rightarrow a_{n+1}\le a_n$ and later “bounded and non-increasing”.

2. Audio
   Observation

   The lecturer says this is “the exact thing that says our sequence is non-decreasing,” but the displayed mathematics and later writing indicate non-increasing.

1. Spoken wording “non-decreasing” is inconsistent with the written inequality direction and the later written conclusion “non-increasing”.

For all $n\ge1$, $a_{n+1}\le a_n$; equivalently, $\{a_n\}$ is non-increasing.

1. $a_n\ge\frac12$ for all $n\ge1$

2. $a_{n+1}-a_n=-\frac23a_n+\frac13$

For every integer $n\ge1$.

1. Formula
   Observation

   Board: “Monotone Convergence $\Rightarrow a_n\to L$.”

2. Audio
   Observation

   “This allows us to use the monotone convergence theorem … implies that the $a_n$ converge to some number $L$.”

Since $\{a_n\}_{n=1}^\infty$ is bounded and monotone (non-increasing), it converges to some real number $L$.

1. $\{a_n\}$ is bounded below

2. $\{a_n\}$ is non-increasing

There exists a real number $L$ such that $a_n\to L$ as $n\to\infty$.

1. Formula
   Observation

   Board solves $L=\frac13(L+1)\Rightarrow \frac23L=\frac13\Rightarrow L=\frac12$.

2. Audio
   Observation

   “Therefore $L$ is equal to one half … our sequence … converges to one half using the monotone convergence theorem.”

If $a_n\to L$, then $L=\frac12$; hence the recursively defined sequence converges to $\frac12$.

1. $a_n\to L$

2. $a_{n+1}=\frac13(a_n+1)$

For the limit $L=\lim_{n\to\infty}a_n$.

1. Formula
   Observation

   The board successively writes $a_2=\frac{1}{3}[1+1]=\frac{2}{3}$ and $a_3=\frac{1}{3}\left[\frac{2}{3}+1\right]=\frac{5}{9}$.

2. Audio
   Observation

   The lecturer narrates each substitution and simplification.

1. Expression

   $$a_2=\frac{1}{3}[1+1]$$
   Explanation

   Substitute the initial value $a_1=1$ into the recurrence $a_{n+1}=\frac{1}{3}(a_n+1)$ with $n=1$.

   Justification

   Direct use of the recursive definition and the given initial condition.

   Shown in the video
2. Expression

   $$a_2=\frac{2}{3}$$
   Explanation

   Simplify the bracket to $2$ and multiply by $\frac{1}{3}$.

   Justification

   Arithmetic simplification.

   Shown in the video
3. Expression

   $$a_3=\frac{1}{3}\left[\frac{2}{3}+1\right]$$
   Explanation

   Substitute $a_2=\frac{2}{3}$ into the recurrence with $n=2$.

   Justification

   Direct use of the recursive definition together with the just-computed value of $a_2$.

   Shown in the video
4. Expression

   $$a_3=\frac{5}{9}$$
   Explanation

   Rewrite $\frac{2}{3}+1=\frac{5}{3}$ and then multiply by $\frac{1}{3}$.

   Justification

   Fraction addition and multiplication.

   Shown in the video

The computed terms are $a_2=\frac{2}{3}$ and $a_3=\frac{5}{9}$, which visually suggest the sequence is moving downward.

1. Formula
   Observation

   Board writes "True if $n=1$" under the claim $a_n\ge \frac{1}{2}$. The earlier board also shows $a_1=1$.

2. Audio
   Observation

   The lecturer says the claim is clearly true and begins the induction proof.

1. The explicit inequality $1\ge \frac{1}{2}$ is not separately written on the board, though it follows immediately from the displayed $a_1=1$.

1. Expression

   $$a_1=1$$
   Explanation

   The initial value of the sequence is given on the board.

   Justification

   Given hypothesis of the recursively defined sequence.

   Shown in the video
2. Expression

   $$1\ge \frac{1}{2}$$
   Explanation

   Therefore the claimed inequality holds when $n=1$.

   Justification

   Elementary comparison of real numbers.

   Derived from the video

The statement $a_n\ge \frac{1}{2}$ is true for $n=1$, so the induction base case holds.

1. Formula
   Observation

   Board writes "Suppose $a_n\ge \frac{1}{2}$" and then the chain $a_{n+1}=\frac{1}{3}(a_n+1)\ge \frac{1}{3}\left(\frac{1}{2}+1\right)=\frac{1}{2}$.

2. Audio
   Observation

   The lecturer explains that by the induction hypothesis the quantity is at least one-third of one-half plus one, then simplifies to one-half.

1. Expression

   $$a_n\ge \frac{1}{2}$$
   Explanation

   Assume the desired inequality holds at index $n$.

   Justification

   Inductive hypothesis.

   Shown in the video
2. Expression

   $$a_{n+1}=\frac{1}{3}(a_n+1)$$
   Explanation

   Write the next term using the recurrence relation.

   Justification

   Definition of the recursively given sequence.

   Shown in the video
3. Expression

   $$a_{n+1}\ge \frac{1}{3}\left(\frac{1}{2}+1\right)$$
   Explanation

   Replace $a_n$ by its lower bound inside the increasing affine expression $\frac{1}{3}(x+1)$.

   Justification

   Monotonicity of addition and multiplication by the positive constant $\frac{1}{3}$, applied to the inductive hypothesis.

   Shown in the video
4. Expression

   $$\frac{1}{3}\left(\frac{1}{2}+1\right)=\frac{1}{3}\cdot \frac{3}{2}$$
   Explanation

   Simplify the sum inside parentheses.

   Justification

   Fraction arithmetic: $\frac{1}{2}+1=\frac{3}{2}$.

   Shown in the video
5. Expression

   $$\frac{1}{3}\cdot \frac{3}{2}=\frac{1}{2}$$
   Explanation

   Multiply the fractions to obtain the final lower bound.

   Justification

   Cancellation in rational arithmetic.

   Shown in the video

If $a_n\ge \frac{1}{2}$, then $a_{n+1}\ge \frac{1}{2}$. Together with the base case, induction proves $a_n\ge \frac{1}{2}$ for all $n\ge 1$.

1. Audio
   Observation

   The lecturer verbally computes $a_{n+1}-a_n$ by substituting the recurrence and combining like terms.

2. Formula
   Observation

   Written result: $a_{n+1}-a_n=\frac13a_n+\frac13-a_n=-\frac23a_n+\frac13$.

1. Expression

   $$a_{n+1}=\frac{1}{3}(a_n+1)$$
   Explanation

   Start from the given recurrence relation.

   Justification

   Given definition of the sequence.

   Shown in the video
2. Expression

   $$a_{n+1}-a_n=\frac{1}{3}(a_n+1)-a_n$$
   Explanation

   Subtract $a_n$ from both sides.

   Justification

   Algebraic manipulation of an equality.

   Shown in the video
3. Expression

   $$=\frac{1}{3}a_n+\frac{1}{3}-a_n$$
   Explanation

   Distribute the factor $\frac13$ across $a_n+1$.

   Justification

   Distributive law.

   Shown in the video
4. Expression

   $$=-\frac{2}{3}a_n+\frac{1}{3}$$
   Explanation

   Combine the $a_n$ terms: $\frac13a_n-a_n=-\frac23a_n$.

   Justification

   Like-term collection.

   Shown in the video

The difference between successive terms is $a_{n+1}-a_n=-\frac23a_n+\frac13$.

1. Formula
   Observation

   Boxed equivalence chain on board: $-\frac23a_n+\frac13\le0\iff \frac13\le\frac23a_n\iff \frac12\le a_n$.

2. Audio
   Observation

   The lecturer explains that showing the expression is $\le0$ is equivalent to $a_n\ge\frac12$, which is already known by induction.

1. Expression

   $$-\frac{2}{3}a_n+\frac{1}{3}\le 0$$
   Explanation

   State the desired inequality for the difference.

   Justification

   Goal is to prove $a_{n+1}-a_n\le0$.

   Shown in the video
2. Expression

   $$\frac{1}{3}\le \frac{2}{3}a_n$$
   Explanation

   Move the $a_n$ term to the other side.

   Justification

   Add $\frac23a_n$ to both sides.

   Shown in the video
3. Expression

   $$\frac{1}{2}\le a_n$$
   Explanation

   Divide by the positive number $\frac23$.

   Justification

   Division by a positive constant preserves inequality direction.

   Shown in the video
4. Expression

   $$a_n\ge \frac{1}{2}$$
   Explanation

   Recognize this as the previously proven lower bound.

   Justification

   Established earlier by induction on the board.

   Shown in the video

Because $a_n\ge\frac12$, it follows that $-\frac23a_n+\frac13\le0$, hence $a_{n+1}-a_n\le0$.

1. Formula
   Observation

   Board writes $L=\frac13(L+1)\Rightarrow \frac23L=\frac13\Rightarrow L=\frac12$.

2. Audio
   Observation

   The lecturer says the left-hand side converges to $L$ and the right-hand side to $\frac13(L+1)$, then solves for $L$.

1. Expression

   $$a_n\to L$$
   Explanation

   Use convergence established by the Monotone Convergence Theorem.

   Justification

   Bounded monotone sequences converge.

   Shown in the video
2. Expression

   $$a_{n+1}\to L$$
   Explanation

   A shifted subsequence of a convergent sequence has the same limit.

   Justification

   Standard property of limits of sequences.

   Derived from the video
3. Expression

   $$L=\frac{1}{3}(L+1)$$
   Explanation

   Pass to the limit in the recurrence $a_{n+1}=\frac13(a_n+1)$.

   Justification

   Continuity of the affine function $x\mapsto \frac13(x+1)$ and limit laws.

   Shown in the video
4. Expression

   $$\frac{2}{3}L=\frac{1}{3}$$
   Explanation

   Multiply out and collect the $L$ terms.

   Justification

   Algebra: $L-\frac13L=\frac13$.

   Shown in the video
5. Expression

   $$L=\frac{1}{2}$$
   Explanation

   Solve for $L$.

   Justification

   Multiply both sides by $\frac32$.

   Shown in the video