How to show sequnce is divergent (math. induction)

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sitia
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Homework Statement



For the sequnce an defined recursively, I have to determine the limiting value, provided that it exists.
a1=2 and a(n+1) = 1/(an)^2 for all n

Homework Equations





The Attempt at a Solution



Ok, so I 've done problems like this but the a(n+1) was biggger than an and so the sequence was increasing and I used mathematical induction and boundness to show it converged. Then I founf the limit.
Now, though I'm having trouble with this sequence becasue it's a decreasing sequence and I know it diverges. How do I go about proving that? What method do I use? At what point is it realized that it diverges?
I'm just really lost :/

Thank you!
 
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sitia said:

Homework Statement



For the sequnce an defined recursively, I have to determine the limiting value, provided that it exists.
a1=2 and a(n+1) = 1/(an)^2 for all n

Homework Equations



The Attempt at a Solution



Ok, so I 've done problems like this but the a(n+1) was biggger than an and so the sequence was increasing and I used mathematical induction and boundness to show it converged. Then I founf the limit.
Now, though I'm having trouble with this sequence becasue it's a decreasing sequence and I know it diverges. How do I go about proving that? What method do I use? At what point is it realized that it diverges?
I'm just really lost :/

Thank you!
Have you written out the first several terms of the sequence?
 
I have...1/4, 1/16...etc
 
sitia said:
Isn't it 1/(2^2)^2
"It", as in a3 is

[itex]\displaystyle a_3=\frac{1}{{a_2}^2}[/itex]
[itex]\displaystyle <br /> =\frac{1}{(1/4)^2}[/itex]

[itex]\displaystyle =\frac{1}{1/16}[/itex]

[itex]= \underline{\ \ ?\ \ }[/itex]
 
sitia said:
Isn't it 1/(2^2)^2

SammyS said:
"It", as in a3 is

[itex]\displaystyle a_3=\frac{1}{{a_2}^2}[/itex]
[itex]\displaystyle <br /> =\frac{1}{(1/4)^2}[/itex]

[itex]\displaystyle =\frac{1}{1/16}[/itex]

[itex]= \underline{\ \ ?\ \ }[/itex]

You shouldn't say "it" unless it is crystal clear what the antecedent is.
 
Oh wow, stupid mistake. Thanks!
So, do I need to show anything to show divergence or just state that it's going to zero and infinity..oscillating, so it's divergent?
 
sitia said:
Oh wow, stupid mistake. Thanks!
So, do I need to show anything to show divergence or just state that it's going to zero and infinity..oscillating, so it's divergent?

Well there certainly are two subsequences that converge to 0 and diverge to infinity.