Convergence of a Sequence


by Icebreaker
Tags: convergence, sequence
Icebreaker
#1
Oct15-05, 03:28 PM
P: n/a
"Let [tex]k\in \mathbb{N}[/tex] and [tex]a_0=k[/tex]. Let [tex]a_n=\sqrt{k+a_{n-1}}, \forall n\geq1[/tex] Prove that [tex]a_n[/tex] converges."

If we look at the similar sequence b_0 = k and b_n = sqrt(a_n-1), then that sequence obviously converges to 1. Unfortunately, b_n<a_n so I can't use the squeeze theorem.

Any hints would be nice.
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1800bigk
1800bigk is offline
#2
Oct15-05, 04:03 PM
P: 42
I would say let lim an = s also lim an-1 would still be s so you can use the limit properties and can get a quadratic with s^2 - s -k=0 you should be able to go from there
devious_
devious_ is offline
#3
Oct16-05, 12:23 AM
P: 347
Quote Quote by 1800bigk
I would say let lim an = s also lim an-1 would still be s so you can use the limit properties and can get a quadratic with s^2 - s -k=0 you should be able to go from there
But you don't know if [itex]\lim a_n[/itex] exists.

Have you tried checking if a_n is monotonic & bounded?

siddharth
siddharth is offline
#4
Oct16-05, 12:39 AM
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Convergence of a Sequence


You should first try to prove that the sequence is bounded.
Then if you show that it monotonically increases or decreases, you can prove that the sequence is convergent.
Icebreaker
#5
Oct16-05, 09:54 AM
P: n/a
It can easily be shown that it's monotonically increasing. However, it's the bounded part that gets me. Maybe I can use Herschfeld's Convergence Theorem?


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