Sequences (Induction?) Problem

Click For Summary
SUMMARY

The discussion focuses on the sequence {an} defined by a1 = sqrt(k) and an+1 = sqrt(k + an) for k > 0. It is established that the sequence is increasing and bounded, leading to the conclusion that the limit as n approaches infinity exists. The limit can be derived from the equation L = sqrt(k + L), which simplifies to L^2 - L - k = 0, allowing for the calculation of the limit using the quadratic formula.

PREREQUISITES
  • Understanding of sequences and limits in calculus
  • Familiarity with mathematical induction techniques
  • Knowledge of solving quadratic equations
  • Basic concepts of monotonicity in sequences
NEXT STEPS
  • Study mathematical induction proofs in detail
  • Learn about monotonic sequences and their properties
  • Explore the quadratic formula and its applications
  • Investigate convergence criteria for sequences
USEFUL FOR

Students in calculus or advanced mathematics courses, educators teaching sequences and limits, and anyone seeking to understand the behavior of recursive sequences.

bz89
Messages
1
Reaction score
0
Homework Statement
Consider the sequence {an} where a1 = sqrt(k), an+1 = sqrt(k + an), and k > 0.

a. Show that {an} is increasing and bounded.

b. Prove that the limit as n approaches infinity of an exists.

c. Find the limit as n approaches infinity of an.

The attempt at a solution

b is straightforward. If you show that an is monotonic and bounded then it has a limit.

I don't really understood how to approach a. The solutions guide suggests some sort of induction that starts with an <= ((1 + sqrt(1 + 4k))/2). I don't understand how I would be able to go from the givens to that point.
 
Physics news on Phys.org
a) calculate a(n+1)-a(n) and draw your conclusion.

b)-

c) In the limit for n tending to infinity you'll get: [tex]L = \sqrt{k+L}[/tex] which you can solve.
 
Last edited:

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K