Finding the Radius and Interval of Convergence of a Series

[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