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Mathematics 14 Online
OpenStudy (simoner):

Calculus 2 question! Help w. power series please

OpenStudy (simoner):

Find the radius and interval of convergence of the series \[\sum_{n=1}^{\infty}\frac{ (-3)^{n} }{ n \sqrt{n}}x^{n}\]

OpenStudy (misty1212):

HI!!

OpenStudy (misty1212):

make a guess ,for the radius of convergence, i bet you are right

OpenStudy (amistre64):

sqrt(3) ??

OpenStudy (misty1212):

if you want to do it step by step we can

OpenStudy (misty1212):

lol \(\sqrt{3}\)

OpenStudy (simoner):

Hi! i tried the problem and got infinity over infinity but that was the wrong answer. the radius of convergence should be 1/3 and interval |x|<3 but idk what i have to do to get the right answer

OpenStudy (misty1212):

we best do it step by step

OpenStudy (simoner):

sorry |x|<1/3

OpenStudy (simoner):

but yes please. thank you

OpenStudy (misty1212):

but one thing for sure, it is more or less obvious that \[\lim_{n\to \infty}\frac{(n+1)\sqrt{n+1}}{n\sqrt{n}}=1\] right ?

OpenStudy (misty1212):

this is where you say "yeah that is pretty obvious"

OpenStudy (simoner):

um why's that obvious?

OpenStudy (misty1212):

think about back in some pre-calc class where you were asked for the "horizontal asympote" of \[f(x)=\frac{(x+1)\sqrt{x+1}}{x\sqrt{x}}\] what would you say?

OpenStudy (simoner):

do you just divide everything by n

OpenStudy (misty1212):

you would say "the degree of the top is the same as the degree of the bottom" (even though they are not really polynomials) so the limit is the ratio of the leading coefficients, namely 1

OpenStudy (simoner):

okay i see now

OpenStudy (misty1212):

you can do it the math teacher way by dividing top and bottom by \(n\) if you like, you still get 1

OpenStudy (michele_laino):

I think that we have to compute this quantity, first: \[\Large L = \mathop {\max \lim }\limits_{n \to + \infty } \sqrt[n]{{\frac{{{3^n}}}{{n\sqrt n }}}}\]

OpenStudy (simoner):

so you would just multiply 3|x| by 1 and this has to be <1

OpenStudy (misty1212):

so now all that is left is \[\frac{3^{n+1}}{3^n}=3\]

OpenStudy (misty1212):

@Michele_Laino i would to with the simple ratio test, not root test

OpenStudy (misty1212):

that means you have to solve \[|3x|<1\] which give you \[|x|<\frac{1}{3}\] or \[-\frac{1}{3}<x<\frac{1}{3}\]

OpenStudy (michele_laino):

it is not the root thest, here we have a power series, so we have to apply a theory slightly different @misty1212

OpenStudy (simoner):

okay thank you so much. so all i have left to do is check endpoints

OpenStudy (michele_laino):

test*

OpenStudy (simoner):

thank you misty1212 i was just stuck on the limit part!

OpenStudy (misty1212):

then you have to do two more things a) replace \(x\) by \(-\frac{1}{3}\) and see what we get, then b) replace \(x\) by \(\frac{1}{3}\) and check that

OpenStudy (misty1212):

ok good \[\color\magenta\heartsuit\]

OpenStudy (michele_laino):

we have: L=3 so from the theory we can write: r=1/L=1/3

OpenStudy (misty1212):

you stuck with something else?

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