به انگلیسی

Proving a sequence settles down

Mathematics · Calculus & Analysis · ages 20-21

Name ______________________ Date ____________

For a_n equal to 3 over n, what is the smallest integer N such that every n greater than or equal to N gives a_n below 0.04? Give N.
Answer: ______________
For a_n equal to 1 over n, what is the smallest integer N such that every n greater than or equal to N gives a_n below 0.01? Give N.
Answer: ______________
Kai says one convergent sequence can have two different limits, 2 and 3. Is Kai right?
Circle one: True False
Suppose a_n tends to 2 and b_n tends to 5. Which step belongs in a proof that a_n plus b_n tends to 7?
- Split epsilon into two halves and take the larger cutoff.
- Check one fixed epsilon only, then reuse that answer for every smaller tolerance.
- Add the two limits first, without choosing cutoffs or bounding an error.
If a_n tends to negative 2 and b_n tends to 3, what does the product law give for a_n b_n?
- negative 6
- negative 2 plus 3, which is 1
- 6
- The product diverges because one limit is negative
What is the limit of a_n equal to (3n plus 1) over (n plus 2)?
- 1 over 2
- It diverges
- 3
What is the limit of a_n equal to 1 over 2 to the n?
- 1
- 0
- 1 over 2
If a_n tends to 4 and b_n tends to negative 1, what is the limit of a_n plus b_n?
- 3
- 5
- negative 3
- 0
For a_n equal to 1 over n, what is the smallest integer N such that every n greater than or equal to N gives a_n below 0.002? Give N.
Answer: ______________
If two sequences converge, taking the larger of their two epsilon cutoffs makes both error bounds hold at once. True or false?
Circle one: True False

Answer key

For grown-ups. Fold this page away before handing over the rest.

Proving a sequence settles down W1-mt_yB5M-DCH8V-s1

76 · Three over n below 0.04 means n is greater than 75, so 76 is the first valid cutoff.
101 · The strict inequality 1 over n below 0.01 means n is greater than 100, so the first valid integer cutoff is 101.
False · Limits are unique: any two candidates must coincide.
Split epsilon into two halves and take the larger cutoff. · Split epsilon into two halves. The two convergence cutoffs then make the triangle inequality bound the sum error below epsilon.
negative 6 · The product law multiplies the limits, so the limit is negative 6.
3 · Divide top and bottom by n and read the leading coefficients.
0 · The terms halve each step and stay within any epsilon past some N.
3 · The sum law adds the two limits, giving 4 plus negative 1 equal to 3.
501 · The strict inequality 1 over n below 0.002 means n is greater than 500, so the first valid integer cutoff is 501.
True · True. Every later index is past both cutoffs, so both bounds hold and the triangle inequality can combine them.