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

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- 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?

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## 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

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