Sequence Convergence/Divergence Question

[Jess Karakov]

Homework Statement

Determine which of the sequences converge or diverge. Find the limit of the convergent sequences.

1) {asubn}= [((n^2) + (-1)^n)] / [(4n^2)]

Homework Equations

[/B]
a1=first term, a2=second term...an= nth term

The Attempt at a Solution

a) So I found the first couple of terms
a1=0
a2=5/16
a3=2/9
a4=17/64

It didn't really look like it converged, but I took the limit as n approached infinity, divided the highest degree in the numerator and the denominator and got the limit to be 1/4. [/B]

 

[member 587159]

Well, what exactly is the problem? The limit of this sequence is 1/4, as you have found yourself. I think you made a mistake in your thought proces though.

Lim(n->∞) [((n^2) + (-1)^n)] / [(4n^2)] = Lim(n->∞) n^2/(4n^2) + Lim(n->∞) (-1)^n/(4n^2) = 1/4 + Lim(n->∞) (-1)^n/(4n^2)
The second of those 2 limits goes to zero, because the numerator (-1)^n = -1, 1, -1, 1, -1, ... and the denominator becomes larger and larger. Only taking the highest degree in numerator and denumerator is not enough because (-1)^n has n in the exponent!

 

[Jess Karakov]

So because the denominator shoots off into infinity and also because (-1)^n oscillates between -1 and 1 forever, the sequence is divergent?

 

[member 587159]

So because the denominator shoots off into infinity and also because (-1)^n oscillates between -1 and 1 forever, the sequence is divergent?

No I meant that Lim(n->∞) (-1)^n/(4n^2) = 0 and therefor: Lim(n->∞) [((n^2) + (-1)^n)] / [(4n^2)] = Lim(n->∞) n^2/(4n^2) + Lim(n->∞) (-1)^n/(4n^2) = 1/4 + 0 = 1/4, so the limit is 1/4.

From what I understand, you found the limit by taking the highest degree in numerator and denumerator, but since there is an n in the exponent, you have to be careful.

 

[Mark44]

Homework Statement

Determine which of the sequences converge or diverge. Find the limit of the convergent sequences.

1) {asubn}= [((n^2) + (-1)^n)] / [(4n^2)]

Homework Equations

[/B]
a1=first term, a2=second term...an= nth term

The Attempt at a Solution

a) So I found the first couple of terms
a1=0
a2=5/16
a3=2/9
a4=17/64

It didn't really look like it converged, but I took the limit as n approached infinity, divided the highest degree in the numerator and the denominator and got the limit to be 1/4. [/B]

The dominant term in the numerator is n², and the dominant term (and only term) in the denominator is 4n². By "dominant" I mean that the larger n gets, the more insignificant the (-1)ⁿ term is. As you say, this sequence converges to 1/4.

You can split this sequence into two sequences: $\{\frac{n^2}{4n^2} \}$ and $\{\frac{(-1)^n}{4n^2} \}$. If you can convince yourself that both sequences converge, then the sum of these two sequences will also converge.

 

[Ray Vickson]

So because the denominator shoots off into infinity and also because (-1)^n oscillates between -1 and 1 forever, the sequence is divergent?

The sequence $s_n = (-1)^n/n^2$ has $s_{10} = 0.01$, $s_{20} = 0.0025$, $s_{1000}= 10^{-6}$, $s_{1000000} = 10^{-12}$, $s_{1000001}= -.999998 \, 10^{-12}$, etc. Does that look to you like a divergent sequence to you?

 

[Jess Karakov]

The sequence $s_n = (-1)^n/n^2$ has $s_{10} = 0.01$, $s_{20} = 0.0025$, $s_{1000}= 10^{-6}$, $s_{1000000} = 10^{-12}$, $s_{1000001}= -.999998 \, 10^{-12}$, etc. Does that look to you like a divergent sequence to you?

No need to be condescending, sir.

 

[Mark44]

So because the denominator shoots off into infinity and also because (-1)^n oscillates between -1 and 1 forever, the sequence is divergent?

The sequence $s_n = (-1)^n/n^2$ has $s_{10} = 0.01$, $s_{20} = 0.0025$, $s_{1000}= 10^{-6}$, $s_{1000000} = 10^{-12}$, $s_{1000001}= -.999998 \, 10^{-12}$, etc. Does that look to you like a divergent sequence to you?

No need to be condescending, sir.

I don't see Ray's response as being condescending. He was responding directly to something you said.