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Convergence of a Sequence

by Icebreaker
Tags: convergence, sequence
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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
#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_
#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
#4
Oct16-05, 12:39 AM
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P: 1,198
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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