radius of convergence and the interval of convergence for the given power series

- Math2400

radius of convergence and the interval of convergence for the given power series

- katieb

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- Math2400

\[\sum_{n=1}^{\infty} \frac{ n ^{2}x ^{2n} }{ n! }\]

- Math2400

my radius is 1. And it think the interval is (-1,1)
Can anyone verify if that's right?

- Zarkon

what test did you use?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- Math2400

ratio

- Zarkon

do it again

- Math2400

ok

- Math2400

wait can the radius be infinity??

- Zarkon

yes

- Math2400

ok..and the interval is (-infin, infin) ?

- Zarkon

yes

- Math2400

lol kk thanks. that's what i got first but i wasn't sure if that was possible haha

- xapproachesinfinity

i got oo as well :)

- Math2400

perfect thanks!! :)

- xapproachesinfinity

limit (x^2 . (n+1/n)^2/(n+1))=0

- Math2400

how about this one if u don't mind: \[\sum_{n=1}^{\infty} \frac{ n^{e} }{ e ^{n} }\] i'm pretty sure the radius is 1 right?

- xapproachesinfinity

that says for any value of x rho is always zero

- Math2400

and then interval is (2,4) or maybe (2,4] not sure, i suck at these

- xapproachesinfinity

i would do root test with that one

- xapproachesinfinity

did you try ratio?

- Zarkon

there is no x in your last problem

- Math2400

yes(: I got the radius to be 1 and then i know the interval is (2,4) but I'm not sure

- xapproachesinfinity

eh hold on you are asking for convergence because there is not x

- xapproachesinfinity

you need x

- Math2400

ohh wait sorry i posted the wrong one. That one is a converging series haha. I meant to post this one: \[\sum_{n=1}^{\infty} \frac{ (x-3)^n }{ \sqrt{n} }\]

- xapproachesinfinity

oh ok
i would go with root test with that too
limit n^1/n=1
rho=|x-3|<1
so raduis is 1
-1 2

- xapproachesinfinity

i guess you got this :)

- Math2400

ok so radius is 1 and then did u check the end points?

- xapproachesinfinity

No i didn't x=4
you have series 1/n^1/2 diverges
x=2
(-1)^n/rootn
i believe this converge by alternating series test

- Math2400

so would it be (2,4] or just (2,4) that's what I'm confused on

- xapproachesinfinity

no converge on 2 not on 4
so [2,4)
otherwise 2<=x<4

- Math2400

ohh haha kk got it :) Thank you so much!

- xapproachesinfinity

no problem :)

- xapproachesinfinity

actually you seem to me good at this, just few more practice and you master it

- Math2400

haha thank you, means a lot @xapproachesinfinity

Looking for something else?

Not the answer you are looking for? Search for more explanations.