# Sequence Convergence/Divergence Question

## 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]

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Math_QED
Homework Helper
2019 Award
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?

Math_QED
Homework Helper
2019 Award
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
Mentor

## 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
Homework Helper
Dearly Missed
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?

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
Mentor
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.