Finding the Radius and Interval of Convergence of a Series

[vela]

Well, take a stab at it proving it and post your attempt.

 

[arl146]

Everything I already said already was my attempt. That's all I got. I just don't get it. Nothing in the book is any different than what I said about my series being less than the one we're comparing it to. I don't know how to show that it's less than. If it even is less than the series we're comparing it to!

 

[arl146]

You know that \sum \frac{1}{n^2} converges (note that the variable is n, not x), but
2) You need to do more than just wave your arms to show convergence. If you are using the comparison test, you need to show that each term of your series is less than the corresponding term of the series you're comparing to. For your problem, \sum \frac{1}{n^2} is a reasonable choice.

like we said, i have the series \frac{n}{n^3+1}

and its from n=1 to infinity .. just doing the first few values, you get 1/2 + 2/9 + 3/28 + ..
which each term is definitely less than the previous term. and just looking at the series you can tell that that will be the case because there's an n with a degree of 1 on top and on bottom there's an n with a degree 3 which shows that. is there another way that I am supposed to show this? if that's right, how do i now prove/show that \frac{n}{n^3+1} is less than or equal to the series \frac{1}{n^2} ? can you do it this way :

ok so you have \frac{n}{n^3+1}. pull out an n on top and bottom and youre left with \frac{1}{n^2+\frac{1}{n}}. this is the same as 1/n^2 except it adds the 1/n on the bottom. which makes my series bigger. if this is right, what does that mean? that its not convergent at x=5? so what now?

 

[vela]

Why does that make your series bigger?

 

[arl146]

because the 1/n adds more to the value than just the 1/n^2 on the bottom. but as n approaches infinity that value doesn't matter so both series are pretty much equal.
is that right or am i thinking about this wrong?

 

[vela]

The 1/n makes the denominator bigger in your series, so what does that imply about the terms of the series compared to 1/n²?

 

[arl146]

yea that's what i said didnt i? OHHHHHH i see what dumb thing i did. yea so the terms of my series is smaller than those of 1/n^2

sorry, i know i do overlook stuff easily like that. i just can't help it, my brain just does it no matter how hard i try

so at x=5 i proved that it is convergent? or is there some how more that I am missing?

 

[arl146]

so.. did i actually prove it this time?

 

[Mark44]

How about writing some mathematics instead of just describing things vaguely with words?

 

[arl146]

ok well

we need \frac{n}{n^3+1} < \frac{1}{n^2} for the test
and that can be proven by:

\frac{n}{n^3+1} = \frac{n(1)}{n(n^2+1/n)} = \frac{1}{n^2+1/n}

so now we are looking at \frac{1}{n^2+1/n} < \frac{1}{n^2} for my series to be convergent since \frac{1}{n^2} converges.

looking at the denominators: n²+\frac{1}{n} is > n²

thus making \frac{n}{n^3+1} = \frac{1}{n^2+1/n} < \frac{1}{n^2} this true

is that good enough?

 

[Mark44]

Looks good!

 

[arl146]

ok great! so can i like write exactly what i have there for my homework?

 

[Mark44]

What you wrote was fine, but a little more verbose and roundabout than necessary. This is how I would say it.

ok well

we need \frac{n}{n^3+1} < \frac{1}{n^2} for the test
and that can be proven by:

\frac{n}{n^3+1} = \frac{n(1)}{n(n^2+1/n)} = \frac{1}{n^2+1/n}

Now, since n² + 1/n > n², for all n >= 1,
then 1/(n² + 1/n) < 1/n², for all n >= 1.

Therefore, n/(n³ + 1) < 1/n², for all n >= 1.
This shows that each term of the series in question is smaller than the corresponding term of the convergent p-series Ʃ1/n².

so now we are looking at \frac{1}{n^2+1/n} < \frac{1}{n^2} for my series to be convergent since \frac{1}{n^2} converges.

looking at the denominators: n²+\frac{1}{n} is > n²

thus making \frac{n}{n^3+1} = \frac{1}{n^2+1/n} < \frac{1}{n^2} this true

is that good enough?

 

[arl146]

ok awesome. that is a lot shorter. and the endpoint check for x=3 is just the same concept so i think i got that. thanks a lot guys!

 

[arl146]

hmm i have another question (hopefully short) ... why do we check the endpoints for convergence? what's the purpose of doing that? do i always do that if the question asks for an interval of convergence and/or radius of convergence?

does checking the endpoints change anything if its convergent or divergent?

 

[Mark44]

The radius of convergence is the same whether the series converges at neither, either, or both endpoints. What does change is the interval of convergence, which could look like (a, b), (a, b], [a, b), or [a, b].

If the question asks for the interval of convergence, you need to check both endpoints.

 

[arl146]

ok yea i just figured that out right as i posted that.. so if they ask just to find the radius of convergence, since that answer comes before finding the interval of convergence, i won't have to check the endpoints since i don't even have to find the interval. correct?

 

[Mark44]

Right