The Epsilon-N Game
A sequence \( (s_n) \) converges to a limit \( L \) if, no matter how small a "target zone" (epsilon, \( \epsilon \)) you pick around \( L \), the sequence eventually enters that zone and never leaves.
How to play:
- Drag the slider to change \( \epsilon \) (the width of the shaded band).
- Watch how the "cutoff point" \( N \) moves to the right as \( \epsilon \) gets smaller.
- For convergence, you must always be able to find such an \( N \), no matter how tiny \( \epsilon \) is.
Formal Definition
A sequence \( (s_n) \) converges to \( L \) if:
\( \forall \epsilon \gt 0, \exists N \in \mathbb{R} \text{ such that } n \gt N \implies |s_n - L| \lt \epsilon \)
Translation:
- \( \forall \epsilon \gt 0 \): For any error margin you challenge me with...
- \( \exists N \): I can find a cutoff point...
- \( n \gt N \implies |s_n - L| \lt \epsilon \): Such that all terms past that point stay within the error margin.
Proof Strategy: Work Backwards
1. Scratch Work (Private)
Start with what you want to be true and solve for \( n \).
// Goal: |1/n - 0| < ε
1/n < ε
1 < nε
n > 1/ε
This tells you what \( N \) should be! Set \( N = 1/\epsilon \).
2. Formal Proof (Public)
Write it forwards, starting with "Let \( \epsilon \gt 0 \)".
Let \( \epsilon \gt 0 \).
Choose \( N = 1/\epsilon \).
Then for any \( n \gt N \), we have \( n \gt 1/\epsilon \).
This implies \( 1/n \lt \epsilon \).
Thus \( |1/n - 0| \lt \epsilon \). Q.E.D.
Practice Problems
True/False
1. A sequence \( (s_n) \) converges to \( L \) if there exists an \( \epsilon \gt 0 \) such that for all \( N \), if \( n \gt N \) then \( |s_n - L| \lt \epsilon \).
2. If a sequence converges, the limit is unique.
3. The value of \( N \) in the definition of a limit must be an integer.
4. If \( |s_n - L| \lt \epsilon \) for all \( n \gt 100 \), then the sequence converges to \( L \).
Fill in the Blank
1. The definition of \( \lim_{n \to \infty} s_n = L \) is: For every \( \epsilon \gt 0 \), there exists an \( N \) such that if \( n \gt N \), then \( |s_n - L| \lt \_\_\_\_\_ \).
2. To prove \( \lim \frac{1}{n} = 0 \), given \( \epsilon \gt 0 \), we should choose \( N = \_\_\_\_\_ \).
3. If a sequence does not converge to any finite limit, we say it \_\_\_\_\_.
4. In the scratch work for proving \( \lim \frac{3n}{n+1} = 3 \), we simplify \( |\frac{3n}{n+1} - 3| \) to get \( \frac{3}{n+1} \). To make this less than \( \epsilon \), we need \( n \gt \_\_\_\_\_ \).
Full Problems
1. Use the definition of the limit to prove that \( \lim_{n \to \infty} \frac{1}{n^2} = 0 \).
Let \( \epsilon \gt 0 \).
We want \( |\frac{1}{n^2} - 0| \lt \epsilon \), which simplifies to \( \frac{1}{n^2} \lt \epsilon \).
Solving for \( n \), we get \( n^2 \gt \frac{1}{\epsilon} \implies n \gt \frac{1}{\sqrt{\epsilon}} \).
Choose \( N = \frac{1}{\sqrt{\epsilon}} \).
Then for all \( n \gt N \), we have \( n \gt \frac{1}{\sqrt{\epsilon}} \implies n^2 \gt \frac{1}{\epsilon} \implies \frac{1}{n^2} \lt \epsilon \). Thus, the limit is 0.
2. Find the limit of the sequence \( s_n = \frac{2n+1}{n+3} \) and prove it using the \( \epsilon-N \) definition.
The limit is 2.
Scratch work: \( |\frac{2n+1}{n+3} - 2| = |\frac{2n+1 - 2(n+3)}{n+3}| = |\frac{-5}{n+3}| = \frac{5}{n+3} \).
We want \( \frac{5}{n+3} \lt \epsilon \implies n+3 \gt \frac{5}{\epsilon} \implies n \gt \frac{5}{\epsilon} - 3 \).
Proof: Let \( \epsilon \gt 0 \). Choose \( N = \frac{5}{\epsilon} - 3 \). For \( n \gt N \), we have \( n \gt \frac{5}{\epsilon} - 3 \implies n+3 \gt \frac{5}{\epsilon} \implies \frac{5}{n+3} \lt \epsilon \). Thus \( |s_n - 2| \lt \epsilon \).
3. Determine if the sequence \( s_n = (-1)^n \) converges or diverges. Prove your answer.
The sequence diverges.
Suppose it converges to \( L \). Let \( \epsilon = 1/2 \). Then there exists \( N \) such that for \( n \gt N \), \( |(-1)^n - L| \lt 1/2 \).
For even \( n \gt N \), \( |1 - L| \lt 1/2 \implies 0.5 \lt L \lt 1.5 \).
For odd \( n \gt N \), \( |-1 - L| \lt 1/2 \implies -1.5 \lt L \lt -0.5 \).
No number \( L \) can satisfy both conditions. Contradiction.
4. Prove that \( \lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0 \).
Let \( \epsilon \gt 0 \). We want \( \frac{1}{\sqrt{n}} \lt \epsilon \implies \sqrt{n} \gt \frac{1}{\epsilon} \implies n \gt \frac{1}{\epsilon^2} \).
Choose \( N = \frac{1}{\epsilon^2} \). Then for \( n \gt N \), \( n \gt \frac{1}{\epsilon^2} \implies \sqrt{n} \gt \frac{1}{\epsilon} \implies \frac{1}{\sqrt{n}} \lt \epsilon \).
5. Given \( \epsilon = 0.1 \), find an \( N \) such that for all \( n \gt N \), \( |\frac{n}{n+1} - 1| \lt \epsilon \).
\( |\frac{n}{n+1} - 1| = |\frac{n - (n+1)}{n+1}| = |\frac{-1}{n+1}| = \frac{1}{n+1} \).
We want \( \frac{1}{n+1} \lt 0.1 \implies n+1 \gt 10 \implies n \gt 9 \).
So any \( N \ge 9 \) works (e.g., \( N=9 \)).