How to show sequnce is divergent (math. induction)

[sitia]

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!

 

[SammyS]

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?

 

[sitia]

I have...1/4, 1/16...etc

 

[Mark44]

I have...1/4, 1/16...etc

a₁ = 2
a₂ = 1/(2²) = 1/4
a₃ = ?
Hint: It's not 1/16.

 

[sitia]

Isn't it 1/(2^2)^2

 

[Mark44]

Isn't it 1/(2^2)^2

Not according to what you wrote in post 1.

a1=2 and a(n+1) = 1/(an)^2 for all n

 

[SammyS]

Isn't it 1/(2^2)^2

"It", as in a₃ is

$\displaystyle a_3=\frac{1}{{a_2}^2}$

$\displaystyle =\frac{1}{(1/4)^2}$

$\displaystyle =\frac{1}{1/16}$

$= \underline{\ \ ?\ \ }$
​

 

[Mark44]

Isn't it 1/(2^2)^2

"It", as in a₃ is

$\displaystyle a_3=\frac{1}{{a_2}^2}$

$\displaystyle =\frac{1}{(1/4)^2}$

$\displaystyle =\frac{1}{1/16}$

$= \underline{\ \ ?\ \ }$
​

You shouldn't say "it" unless it is crystal clear what the antecedent is.

 

[sitia]

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?

 

[happysauce]

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.