Calc 2: Power Series - Find Radius & Interval of Convergence

[arl146]

Homework Statement

Find the radius of convergence and interval of convergence of the series:
[itex]\sum[/itex] (-1)ⁿ * ( x^(2n) / (2n)! )
for n=0 to infinity

Homework Equations

well, any of the divergence tests: ratio, limit comparison, etc.

The Attempt at a Solution

I don't get power series at all ... well, series at all. All I know is you have to use one of the divergence tests for this .. but I don't know really how to apply them to problems .. any help at all would be appreciated

but, i was thinking to use the ratio test. not sure if that's the best way to go.. can someone also explain how you know which test to use??

So anyways, if you do the ratio test, I ended up with, [ (-1)*(x^2)*(2n)! ] / (2n+2)!
and now I don't know what else to do with this. i see that x doesn't depend on n so pull that out of the limit .. and that's as far as i get

 

[Mark44]

Homework Statement

Find the radius of convergence and interval of convergence of the series:
[itex]\sum[/itex] (-1)ⁿ * ( x^(2n) / (2n)! )
for n=0 to infinity

Homework Equations

well, any of the divergence tests: ratio, limit comparison, etc.

The Attempt at a Solution

I don't get power series at all ... well, series at all. All I know is you have to use one of the divergence tests for this .. but I don't know really how to apply them to problems .. any help at all would be appreciated

but, i was thinking to use the ratio test. not sure if that's the best way to go.. can someone also explain how you know which test to use??

So anyways, if you do the ratio test, I ended up with, [ (-1)*(x^2)*(2n)! ] / (2n+2)!
and now I don't know what else to do with this. i see that x doesn't depend on n so pull that out of the limit .. and that's as far as i get

In the ratio test, you're dealing with the absolute values of terms, so you can ignore the (-1)ⁿ factor.

That leaves you with |x|² * (2n)! / (2n + 2)!

(2n + 2)! = (2n + 2) * (2n + 1) * (2n)!, right?

Under what conditions do you get convergence with the Ratio Test?

 

[arl146]

yea that makes sense. so you always ignore the (-1)^n if using ratio test??

so, if the limit L < 1 it is convergent and if it is > 1 then it is divergent. how do we know if it is greater or less than 1?

 

[arl146]

doesnt that just leave with like x = + or - 1 ? or am i going about that wrong

 

[HallsofIvy]

The ratio test says that a power series converges (absolutely) if that ratio is less than one. Your condition is not that x be equal to 1 or -1 but that it lie between -1 and 1. Now, what is the interval of convergence and radius of convergence?

 

[arl146]

oops yea that's what i meant, x is between -1 and 1. so the radius of convergence is just 1? and the interval of convergence is (-1,1) ? ugh i don't know

 

[Mark44]

I think you need to back up a minute. You're looking at this limit:
[tex]\lim_{n \to \infty} \frac{|x|^2}{(2n + 2)(2n + 1)}[/tex]

For convergence, this limit needs to be < 1. For what x will that be true?

 

[arl146]

as long as x > 1 ?

 

[Mark44]

I think you're guessing. Since x doesn't have anything to do with n, you can move it outside the limit. If you do that, what do you get for the limit?

 

[arl146]

youd have 1/(2n+2)(2n+1) and when n goes to infinity, that'd be 1/infinity so isn't that just like 0 ? ugh calc 3 was so much easier than this =/ that answer doesn't make sense

 

[arl146]

wait no ... when n goes to 0 ... that'd be 1/2 right? so x would have to be less than + or - sqrt(2) ?

 

[arl146]

Ok so I just went through this problem again and I get stuck with this:

x*(lim as n approaches infinity of 1/((2n+2)*(2n+1)))

So doesn't that limit just go to 0? If that's true, where do I go from there to find the interval of convergence and radius of convergence ?

 

[Mark44]

wait no ... when n goes to 0 ... that'd be 1/2 right? so x would have to be less than + or - sqrt(2) ?

The limit is NOT as n goes to 0, so that shouldn't even be considered.

 

[arl146]

That's why I posted again, I realized that. So is the post after that in which it goes to infinity correct?

 

[Mark44]

Ok so I just went through this problem again and I get stuck with this:

x*(lim as n approaches infinity of 1/((2n+2)*(2n+1)))

Actually, it's
[tex]|x| \lim_{n \to \infty}\frac{1}{(2n + 2)(2n + 1)}[/tex]

And yes, this limit is 0.

So doesn't that limit just go to 0? If that's true, where do I go from there to find the interval of convergence and radius of convergence ?

Since the limit above is zero, no matter what x is, what does that say about any restrictions on x? That is related to what the interval of and radius of convergence are.

 

[arl146]

Isn't that what I have I just didn't have the absolute value signs? I'm on my phone I keep forgetting them. But yea so for all x, it goes to zero so R= Infinity right? So that's the radius of convergence and the Interval is just (-infinity, infinity) ?

 

[Mark44]

Isn't that what I have I just didn't have the absolute value signs? I'm on my phone I keep forgetting them.

Don't forget them - they're important.

But yea so for all x, it goes to zero so R= Infinity right? So that's the radius of convergence and the Interval is just (-infinity, infinity) ?

Right.

 

[arl146]

Yayyy thanks so much!