Limit Logic Slides Limit Logic
The \(\epsilon\text{-}N\) Definition of Convergence
The Paradox of 0.999...
Does the sequence \((0.9, 0.99, 0.999, ...)\) ever actually reach 1?
Intuitively, we say the limit is 1. But in Real Analysis, "intuition" is where logic goes to die.
0.999... = 1 ?
Today we build the machinery to prove this beyond a shadow of a doubt.
Refining "Close"
"The terms \(a_n\) get closer and closer to \(L\)." Problem: This could mean \(1/n\) gets "closer" to -1, but never arrives.
"The distance \(|a_n - L|\) eventually stays as small as we want." Better: We need a way to quantify "as small as we want" and "eventually".
Formal Definition
A sequence \((a_n)\) converges to a real number \(L\) if:
\(\forall \epsilon > 0, \exists N \in \mathbb{N}\) such that
The Proof Blueprint
The Challenge
"Let \(\epsilon > 0\) be given."
We don't pick \(\epsilon\). Someone else does, and it's tiny.
Scratch Work
Solve for \(n\).
Manipulate \(|a_n - L| < \epsilon\) to isolate \(n\). This reveals our \(N\).
The Proof
"Choose \(N = ...\)"
Show that for this \(N\), the implication holds true.
Example Case
Prove \(\lim_{n \to \infty} \frac{1}{n} = 0\)
Given: \(\epsilon > 0\)Want: \(|\frac{1}{n} - 0| < \epsilon \iff \frac{1}{n} < \epsilon\)Solve: \(n > \frac{1}{\epsilon}\)Choose: \(N = \lceil \frac{1}{\epsilon} \rceil\) (Archimedean Property)Verify: If \(n > N\), then \(n > \frac{1}{\epsilon} \implies \frac{1}{n} < \epsilon\). Q.E.D.
Workshop Time
Transition to your worksheets. We will be constructing rigorous proofs for arithmetic and geometric sequences. Remember: \(\epsilon\) is your only constraint.
3
Problems
20
Minutes
Epsilon-N Workshop Worksheet Epsilon-N Workshop
Topic: Rigorous Limit Proofs for Real Sequences
Lesson 01
Formalizing the Limit
Scholar Name
Date
Primary Objective
Utilize the formal definition of convergence to construct rigorous \(\epsilon\text{-}N\) proofs for given real-valued sequences. Focus on the logical flow: (1) Identify target inequality, (2) Perform scratch work to solve for \(n\), (3) Formally define \(N\).
Core Definition
\(\lim_{n \to \infty} a_n = L \iff \forall \epsilon > 0, \exists N \in \mathbb{N} \text{ s.t. } n > N \implies |a_n - L| < \epsilon\)
1
Arithmetic Sequence Decay
Prove that the sequence \(a_n = \frac{3n + 1}{n + 2}\) converges to \(L = 3\).
Phase I: Scratch Work
Phase II: Formal Proof
2
Geometric Sequence Convergence
Prove that the sequence \(x_n = r^n\) where \(|r| < 1\) converges to \(0\). (Hint: Use the fact that \(\frac{1}{|r|} = 1 + h\) for some \(h > 0\) and apply Bernoulli's Inequality or logarithms.)
Formal Proof Construction
Conceptual Reflection
Explain why we cannot simply choose \(N = \frac{1}{\epsilon}\) if the scratch work yields \(n > \frac{5}{\epsilon}\). Why must the relationship between \(n\) and \(N\) be strictly maintained?
Convergence Proof Guide Teacher Resource Convergence Proof Guide
Teacher Resource | Lesson 01
Instructional Strategy
The Epsilon Struggle
Graduate students often struggle with the transition from "solving for x" to "proving for all epsilon." Emphasize that we are building a machine: if you give me an epsilon, my N will produce a tail of the sequence that satisfies the distance condition.
Common Misconception
Students often think they need to find the "best" or "smallest" N. Clarify that any N that works is sufficient, but finding a simple N (often by overestimating terms in the scratch work) is preferred for clarity in proofs.
Problem 1: \(a_n = \frac{3n + 1}{n + 2} \to 3\)
Scratch Work logic
Goal: Find \(N\) such that \(n > N \implies |\frac{3n+1}{n+2} - 3| < \epsilon\)
\(|\frac{3n+1 - 3(n+2)}{n+2}| = |\frac{3n+1 - 3n - 6}{n+2}| = |\frac{-5}{n+2}| = \frac{5}{n+2}\)
We want \(\frac{5}{n+2} < \epsilon \iff n+2 > \frac{5}{\epsilon} \iff n > \frac{5}{\epsilon} - 2\)
Formal Proof Structure
"Let \(\epsilon > 0\). Choose \(N \in \mathbb{N}\) such that \(N > \frac{5}{\epsilon} - 2\). Then for all \(n > N\), we have:
Problem 2: \(r^n \to 0\) for \(|r| < 1\)
Strategic Approaches
Approach A (Logarithms): Solving \(|r|^n < \epsilon\) by taking \(\ln\) on both sides. Note that \(\ln|r|\) is negative, so the inequality flips: \(n > \frac{\ln \epsilon}{\ln |r|}\).Approach B (Bernoulli's): Let \(\frac{1}{|r|} = 1 + h\) where \(h > 0\). Then \(\frac{1}{|r|^n} = (1+h)^n \ge 1 + nh > nh\). Thus \(|r|^n < \frac{1}{nh}\). Setting \(\frac{1}{nh} < \epsilon\) gives \(n > \frac{1}{h\epsilon}\). This is often considered more "purely algebraic."
Grading Tip
Check specifically for the quantifier order. A student who writes "There exists \(N\) for all \(\epsilon\)" has fundamentally failed the logic of the proof, even if their algebra is correct.
Monotone Convergence Slides Monotone Trends
Boundedness and the MCT
Perpetual Growth
Consider a sequence where every term is larger than the one before it.
Does "always increasing" imply "divergence to infinity"?
\(a_1 \le a_2 \le a_3 \le ...\)
If it stays below a ceiling, it must settle somewhere.
Analytic Constraints
Boundedness
A sequence \((a_n)\) is bounded if there exists \(M > 0\) such that \(|a_n| \le M\) for all \(n \in \mathbb{N}\).
Monotonicity
A sequence is monotone if it is either increasing (\(a_n \le a_{n+1}\)) or decreasing (\(a_n \ge a_{n+1}\)).
Theorem: MCT
Monotone Convergence Theorem
"Every bounded monotone sequence of real numbers is convergent."
This is a fundamental consequence of the Completeness Axiom of \(\mathbb{R}\).
The Intuition of L.U.B.
If a sequence is increasing and bounded above, the limit is simply the Supremum of the set of its terms.
\(\lim_{n \to \infty} a_n = \sup \{a_n : n \in \mathbb{N}\}\)
Proof Sketch: For any \(\epsilon\), \(\sup - \epsilon\) is not an upper bound, so at least one term \(a_N\) exceeds it. Monotonicity ensures all subsequent terms stay within that \(\epsilon\)-window.
Recursive Analysis
Consider \(a_{n+1} = \sqrt{2 + a_n}\) with \(a_1 = \sqrt{2}\).
Step 1: Induction
Prove the sequence is increasing and bounded above by 2.
Step 2: Limit Hunt
Once MCT guarantees a limit \(L\), solve \(L = \sqrt{2+L}\).
Recursive Riddles
In today's workshop, you will apply the MCT to sequences defined recursively. You must establish both monotonicity and boundedness before calculating any limits.
Induction Proofs
Limit Solves
Recursive Riddles Worksheet Recursive Riddles
Topic: Applying the Monotone Convergence Theorem (MCT)
Lesson 02
Boundedness & Monotonicity
Scholar Name
Date
The Standard Procedure for Recursive Sequences
1. Bound Use induction to show \(|a_n| \le M\) for all \(n\).
2. Monotone Show \(a_{n+1} - a_n\) stays consistently non-negative or non-positive.
3. Solve Set \(L = f(L)\) and solve for the limit value.
1
The Square Root Spiral
Define a sequence by \(a_1 = \sqrt{3}\) and \(a_{n+1} = \sqrt{3 + a_n}\) for \(n \ge 1\).
Part A: Induction Proof (Boundedness)
Show that \(0 < a_n < 3\) for all \(n\).
Part B: Monotonicity
Show that \(a_n < a_{n+1}\) for all \(n\).
Part C: Limit Conclusion
Justify the existence of the limit and find its value.
2
Newton's Method Sequence
Let \(x_1 > \sqrt{2}\). Define \(x_{n+1} = \frac{1}{2} (x_n + \frac{2}{x_n})\). This is the ancient Greek method for approximating \(\sqrt{2}\).
Arithmetic-Geometric Mean Inequality
Show that \(x_n \ge \sqrt{2}\) for all \(n \ge 2\). (Hint: \(x_{n+1} - \sqrt{2} = \frac{(x_n - \sqrt{2})^2}{2x_n}\))
Monotonicity Analysis
Determine if the sequence is increasing or decreasing for \(n \ge 2\). Provide proof.
Final Proof
Combine your findings to prove \(\lim_{n \to \infty} x_n = \sqrt{2}\).
MCT Mastery Key Teacher Resource MCT Mastery Key
Teacher Solution & Facilitation Guide | Lesson 02
1
The Square Root Spiral Solution
Boundedness Induction
Base: \(a_1 = \sqrt{3} < 3\)Hypothesis: \(a_k < 3\)Step: \(a_{k+1} = \sqrt{3 + a_k} < \sqrt{3 + 3} = \sqrt{6} < 3\). Done.
Monotonicity
Check \(a_{n+1}^2 - a_n^2 = (3 + a_n) - a_n^2 = -(a_n^2 - a_n - 3)\). Roots are \(\frac{1 \pm \sqrt{13}}{2}\). Since \(1 < a_n < 3\), the quadratic is negative, so \(a_{n+1}^2 > a_n^2\).
Limit Calculation
\(L = \sqrt{3 + L} \implies L^2 - L - 3 = 0 \implies L = \frac{1 + \sqrt{13}}{2} \approx 2.30\)
Note: Reject negative root as \(a_n > 0\).
2
Newton's Method Solution
Key Logic: The Gap Proof
The hint \(x_{n+1} - \sqrt{2} = \frac{(x_n - \sqrt{2})^2}{2x_n}\) is crucial. Since the right side is always non-negative for \(x_n > 0\), we have \(x_{n+1} \ge \sqrt{2}\) for all \(n\). This establishes the Lower Bound.
Monotonicity via Ratio/Difference
\(x_{n+1} - x_n = \frac{1}{2}(x_n + \frac{2}{x_n}) - x_n = \frac{1}{x_n} - \frac{1}{2}x_n = \frac{2 - x_n^2}{2x_n}\).
Since \(x_n^2 \ge 2\) for all \(n \ge 2\), the numerator is \(\le 0\). Thus the sequence is decreasing .
Pedagogical Warning
Many students will skip proving existence and go straight to \(L = \frac{1}{2}(L + 2/L)\). Explicitly penalize solutions that don't cite the MCT. Calculating a limit that doesn't exist (e.g., \(a_{n+1} = -a_n\)) is a classic analyst's trap.
Cauchy Closeness Slides Internal Closeness
Cauchy Sequences & Completeness
The Problem with Limits
To prove a sequence converges via \(\epsilon\text{-}N\), you must know the limit \(L\) beforehand.
What if you don't know where the sequence is going, but you know the terms are crowding together?
"Internal convergence" vs "External convergence"
Formal Definition
A sequence \((a_n)\) is a Cauchy Sequence if:
\(\forall \epsilon > 0, \exists N \in \mathbb{N}\) such that
"The terms get and stay arbitrarily close to each other."
The Crucial Link
Theorem 1
Every convergent sequence is Cauchy.
Proof: Use the triangle inequality on \(|a_m - a_n| = |(a_m - L) + (L - a_n)|\).
The Big Question
Is every Cauchy sequence convergent?
The answer depends entirely on the world (set) you live in.
The Failure of Rationals (\(\mathbb{Q}\))
In \(\mathbb{Q}\), the sequence of rational approximations of \(\sqrt{2}\) is Cauchy...
...but it has no limit in \(\mathbb{Q}\).
The rational "line" is full of holes. Cauchy sequences can fall right through them.
\(\{1, 1.4, 1.41, 1.414, ...\}\)
The Completeness Axiom
"In the Real Numbers (\(\mathbb{R}\)), a sequence converges if and only if it is a Cauchy sequence."
Density
No Gaps
Convergence
Can you arrive without a destination?
If a hiker walks and each step is half the size of the previous one, do they always arrive at a point? Contrast the rational trail vs. the real road.
Next: Investigating Q vs R Workshop
Completeness Comparison Worksheet The Completeness Gap
Topic: Analyzing \(\mathbb{Q}\) vs \(\mathbb{R}\) via Cauchy Sequences
Lesson 03
Cauchy & Completeness
Analyst Name
Timestamp
1
Proving Internal Convergence
Prove that if a sequence \((a_n)\) is convergent to \(L\), then it is necessarily a Cauchy sequence.
Logic Breakdown
Start by choosing an \(\epsilon > 0\). Use the fact that there exists an \(N\) such that \(n > N \implies |a_n - L| < \epsilon/2\). Then apply the Triangle Inequality to the expression \(|a_m - a_n|\).
2
The Hole in \(\mathbb{Q}\)
Consider the sequence \((q_n)\) of rational numbers defined by: \[q_1 = 1, \quad q_{n+1} = \frac{q_n}{2} + \frac{1}{q_n}\] It can be shown that \(q_n^2 \to 2\). Prove that \((q_n)\) is a Cauchy sequence in \(\mathbb{Q}\), but its limit does not exist in \(\mathbb{Q}\).
Step A: Justify Cauchy Status
Why does the fact that it converges in R imply it is Cauchy in Q?
Step B: Contradiction Proof
Suppose \(\lim q_n = p/q \in \mathbb{Q}\). Use the algebraic definition of the sequence to reach a contradiction (showing \(\sqrt{2} \notin \mathbb{Q}\)).
3
Completeness in Subsets
Is the open interval \((0, 1)\) complete as a metric space under the standard absolute value metric? Justify your answer using a specific sequence.
The Big Picture
Completeness is the property that "Cauchy implies Convergent." If a space is complete, you can prove things exist (limits, solutions to equations) simply by showing the elements stay close together. Why is this property essentially for Calculus to work?
Metric Completeness Key Teacher Resource Completeness Key
Teacher Solution Guide | Lesson 03
Problem 1: Convergent \(\implies\) Cauchy
Proof Strategy
|a_m - a_n| = |(a_m - L) + (L - a_n)| \le |a_m - L| + |a_n - L|
Since \(a_k \to L\), for any \(\epsilon > 0\), choose \(N\) such that \(k > N \implies |a_k - L| < \epsilon/2\). Then for any \(m, n > N\), the triangle inequality yields \(|a_m - a_n| < \epsilon/2 + \epsilon/2 = \epsilon\).
Problem 2: The Hole in \(\mathbb{Q}\)
Step A: The sequence is defined in \(\mathbb{Q}\) and converges to \(\sqrt{2}\) in \(\mathbb{R}\). Since every convergent sequence in \(\mathbb{R}\) is Cauchy, and the terms are in \(\mathbb{Q}\), the sequence is Cauchy in \(\mathbb{Q}\) (the distances between terms are the same).
Step B: If \(\lim q_n = L \in \mathbb{Q}\), then \(L = L/2 + 1/L \implies L^2/2 = 1 \implies L^2 = 2\). But there is no rational number whose square is 2. Thus the limit does not exist in \(\mathbb{Q}\).
Problem 3: \((0, 1)\) is not complete
Counter-example: Consider the sequence \(a_n = 1/n\). Each term is in \((0, 1)\). The sequence is Cauchy because it converges in \(\mathbb{R}\) (to 0). However, the limit 0 is not in the space \((0, 1)\). Therefore, the space is not complete.
Clarify for students: Completeness is a property of the space and the metric, not just the numbers themselves.
BW Theorem Slides Finding Order
Subsequences & Bolzano-Weierstrass
Order from Chaos
Consider the sequence \(a_n = (-1)^n\). It bounces between -1 and 1 forever. It clearly does not converge.
But if we look only at the even terms, we see a perfectly stable, constant sequence of 1s.
-1, 1, -1, 1, -1, 1...
Defining the Selection
Let \((a_n)\) be a sequence. A subsequence is a sequence of the form \((a_{n_k})\) where:
n_1 < n_2 < n_3 < ... < n_k < ...
"We pick terms in their original order, but we can skip as many as we want."
The Convergence Bridge
T1
If \((a_n) \to L\), then every subsequence \((a_{n_k}) \to L\).
T2
If \((a_n)\) has two subsequences converging to different limits, then \((a_n)\) diverges .
Bolzano-Weierstrass Theorem
"Every bounded sequence of real numbers has a convergent subsequence."
This is a profound statement about Compactness . No matter how much a bounded sequence "jitters," it must cluster somewhere.
Accumulation Points
A number \(x\) is an accumulation point (or cluster point) of \((a_n)\) if there exists a subsequence \((a_{n_k})\) such that \(a_{n_k} \to x\).
Case Study
Sequence: \(a_n = \sin(n)\)
By BW, since \(\sin(n)\) is bounded by [-1, 1], it MUST have at least one accumulation point. In fact, every point in [-1, 1] is an accumulation point.
Searching for Subsequences
Your task is to identify accumulation points for oscillating sequences and apply the Bolzano-Weierstrass theorem to prove existence where visual intuition fails.
PROCEED TO WORKSHEET 04
Subsequence Search Worksheet Subsequence Search
Topic: Accumulation Points & Bolzano-Weierstrass
Lesson 04
Subsequences & BW
Investigator
Case Date
1
The Oscillating Target
Identify all accumulation points for the sequence \(a_n = \sin\left(\frac{n\pi}{2}\right) + \frac{1}{n}\). For each accumulation point found, explicitly define the subsequence indices \(n_k\) that converge to it.
Evidence Analysis
Convergence Proofs
2
Existence without Construction
Let \((x_n)\) be a sequence such that \(0 \le x_n \le 10\) for all \(n\). Suppose that for every convergent subsequence \((x_{n_k})\), the limit \(L\) is always greater than 5.
Prove: There exists an \(N\) such that \(x_n > 4\) for all \(n > N\). (Hint: Use proof by contradiction and the Bolzano-Weierstrass theorem.)
3
The Enumeration of \(\mathbb{Q} \cap [0,1]\)
Let \((r_n)\) be an enumeration of all rational numbers in the interval \([0, 1]\). Prove that every real number \(x \in [0, 1]\) is an accumulation point of this sequence.
Compactness Guide Teacher Resource Subsequence Solutions
Teacher Solution & Facilitation Guide | Lesson 04
Problem 1: Accumulation Points for \(\sin(n\pi/2) + 1/n\)
Point: 1
Choose \(n_k = 4k + 1\)
\(\sin((4k+1)\pi/2) = 1\)
Point: -1
Choose \(n_k = 4k + 3\)
\(\sin((4k+3)\pi/2) = -1\)
Point: 0
Choose \(n_k = 2k\)
\(\sin(2k\pi/2) = 0\)
Problem 2: Proof via BW
Proof by Contradiction: Suppose no such \(N\) exists. Then for all \(k\), there exists \(n_k > k\) such that \(x_{n_k} \le 4\). This defines a subsequence \((x_{n_k})\). Since the original sequence is bounded in \([0, 10]\), the subsequence \((x_{n_k})\) is also bounded. By BW, \((x_{n_k})\) has a convergent sub-subsequence \((x_{n_{k_j}}) \to L\). Since all terms \(x_{n_{k_j}} \le 4\), the limit \(L \le 4\). But this contradicts the premise that every convergent subsequence has a limit \(L > 5\).
Problem 3: Rational Density
Let \(x \in [0, 1]\). By the Density of Rationals , for every \(k \in \mathbb{N}\), there exists a rational \(q \in \mathbb{Q} \cap [0, 1]\) such that \(|x - q| < 1/k\). Since \((r_n)\) is an enumeration of all such rationals, we can pick indices \(n_k\) such that \(|x - r_{n_k}| < 1/k\) and \(n_k > n_{k-1}\). This subsequence \((r_{n_k})\) converges to \(x\).
Asymptotic Oscillation Slides Asymptotic Bounds
Limit Superior & Limit Inferior
The Limit that Isn't
We know \(a_n = (-1)^n(1 + 1/n)\) diverges. It jumps between "nearly 1" and "nearly -1".
But it clearly has a "top" and a "bottom" behavior as \(n \to \infty\).
Sequence Behavior
1.1, -1.05, 1.03, -1.02...
Watching the Tail
Let \(b_N = \sup \{a_n : n \ge N\}\).
As \(N\) increases, we are looking at smaller and smaller "tails" of the sequence.
Observation:
Since we are taking the supremum of smaller sets, \(b_N\) is a decreasing sequence. By MCT (or generalized limit), it must have a limit.
Limit Superior
\(\limsup a_n\)
The limit of the "tail supremums". It is the largest accumulation point of the sequence.
Limit Inferior
\(\liminf a_n\)
The limit of the "tail infimums". It is the smallest accumulation point of the sequence.
The Ultimate Convergence Test
A sequence \((a_n)\) converges to \(L\) if and only if:
\(\liminf a_n = \limsup a_n = L\)
"When the top and bottom squeeze together, the whole sequence must follow."
Asymptotic Properties
\(\liminf a_n \le \limsup a_n\) always holds.
If \(a_n \le b_n\), then \(\limsup a_n \le \limsup b_n\).
Useful in Analysis when a limit might not strictly exist.
Handling Oscillation
We will now calculate limits superior and inferior for sequences that refuse to settle. This is the final tool in your sequence analysis toolkit.
INF
Floor
SUP
Ceiling
Oscillation Analysis Worksheet Oscillation Analysis
Topic: Limits Superior and Inferior
Lesson 05
Lim Sup & Lim Inf
Lead Analyst
Analysis Date
1
Calculating Asymptotic Levels
Calculate the limit superior and limit inferior for the following sequence: \[a_n = \left(1 + \frac{1}{n}\right)\cos(n\pi)\]
Tail Supremum Analysis
Tail Infimum Analysis
Final \(\limsup\)
Final \(\liminf\)
2
Properties of Addition
Let \((a_n)\) and \((b_n)\) be bounded sequences.
Prove: \(\limsup (a_n + b_n) \le \limsup a_n + \limsup b_n\).
3
Asymptotic Squeeze
Prove that a bounded sequence \((a_n)\) converges to \(L\) if and only if \(\liminf a_n = \limsup a_n = L\). You may assume the Bolzano-Weierstrass theorem in your proof.
Asymptotic Behavior Key Teacher Resource Asymptotic Logic Key
Teacher Solution Guide | Lesson 05
1
Calculating (1+1/n)cos(npi)
The sequence oscillates: \(a_n = 1 + 1/n\) for even \(n\), and \(a_n = -(1 + 1/n)\) for odd \(n\).
Supremum Analysis
\(b_N = \sup \{a_n : n \ge N\}\). For any \(N\), the set contains values like \(1 + 1/k\) where \(k\) is even and \(k \ge N\). The supremum of this tail is \(1 + 1/k_{min}\). As \(N \to \infty\), \(1+1/k \to 1\). Thus \(\limsup a_n = 1\).
Infimum Analysis
\(c_N = \inf \{a_n : n \ge N\}\). The tail contains \(-(1 + 1/k)\). The infimum is the "most negative" value. As \(N \to \infty\), \(-(1+1/k) \to -1\). Thus \(\liminf a_n = -1\).
2
Sum Inequality Proof
Proof: For any \(n \ge N\), \(a_n + b_n \le \sup_{k \ge N} a_k + \sup_{k \ge N} b_k\). Taking the supremum of the left side over \(n \ge N\) maintains the inequality: \(\sup_{n \ge N} (a_n + b_n) \le \sup_{k \ge N} a_k + \sup_{k \ge N} b_k\). Taking the limit as \(N \to \infty\) on both sides gives \(\limsup (a_n+b_n) \le \limsup a_n + \limsup b_n\).
Example of < : Let \(a_n = (-1)^n\) and \(b_n = (-1)^{n+1}\). Then \(a_n + b_n = 0\), so \(\limsup(a_n+b_n) = 0\). However, \(\limsup a_n = 1\) and \(\limsup b_n = 1\), so \(1+1 = 2\). Clearly \(0 < 2\).
3
Convergence Criterion
Forward: If \(a_n \to L\), then every subsequence converges to \(L\). Since \(\limsup\) and \(\liminf\) are the largest and smallest accumulation points (limits of subsequences), they must both equal \(L\).
Reverse: If \(\liminf a_n = \limsup a_n = L\), then for any \(\epsilon\), eventually the tail infimum is \(> L-\epsilon\) and the tail supremum is \(< L+\epsilon\). Thus \(L-\epsilon < a_n < L+\epsilon\) for all large \(n\), satisfying the definition of convergence.
