radius and interval of convergence of the series sqrt(n)*x^n as n goes from 1 to infinity?? please help..

- anonymous

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- anonymous

the radius = 0, since x stands alone here :)

- anonymous

wait

- anonymous

Just give me a minute.

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- anonymous

ok.

- anonymous

are you sure it's the sqrt of n and not the power of n?

- anonymous

yes

- anonymous

Well.. I am going to use the ratio test first to find the radius of convergence.

- anonymous

you can't use the ratio test here anwar, since it's to the power of 1/n and not n, you'll have to use the root test here ^_^ and hey :)

- anonymous

but since the question doesn't have any sort of exponential function, then "maybe" you can :)

- anonymous

but it's going to be difficult

- anonymous

Yes I can, and it won't be difficult. And hello! :)

- anonymous

lol, root test is much easier here :)

- anonymous

let:
\[an = x^n\]
so \[|an^{\frac{1}{n}} |= (|x^n)^{\frac{1}{n}} |= |x|\]

- anonymous

and :\[0\le |x| < 1\]
so the radius is = 1 in this case, I guess, correct me if I'm wrong please :)

- anonymous

so since :\[0 \le |x| < 1\]
then :
\[-1 < x < 1\] is the interval of convergence ^_^

- anonymous

I took this section yesterday morning during my calculus class lol

- anonymous

so I might have a couple of mistakes ^_^" maybe

- anonymous

just so its clear the series is..\[\sqrt{n}*x ^{n}\]

- anonymous

oh >_

- anonymous

\[\left| {a_{n+1} \over a_n} \right|=\left| {\sqrt{n+1}. x^{n+1} \over \sqrt{n} x^n} \right|=\left| x \right|.\left| {\sqrt{n+1} \over \sqrt n} \right|\]

- anonymous

then find the limit , oh my bad, I completely misunderstood the question >_<

- anonymous

proceed with anwar's answer, it seems correct :)

- anonymous

oh sorry for the confusion..ok yea i got till there..and then?

- anonymous

continue anwar, you're on the right track :)

- anonymous

So now take the limit as n goes to infinity:
\[\lim_{n \rightarrow \infty}{\left| x \right|.\left| {\sqrt{n+1} \over \sqrt n} \right|}=\left| x \right|\]

- anonymous

oh, almost the same LOL =P bleh

- anonymous

Is the limit part clear?

- anonymous

yess

- anonymous

oh congrats on being a champion anwar ^_^

- anonymous

Good. As I said before, the series is convergent when the limit value is <1. In our case, the radius of convergent is:
\[R=\left| x \right|<1 \implies -1

- anonymous

Thanks sstarica :)

- anonymous

:) sure

- anonymous

the radius of convergence = 1 and the interval is the one on top lol ^_^

- anonymous

Now you should check the endpoints. check if the series converges at x=1 and x=-1.

- anonymous

but i thought it had to be less than one to converge..

- anonymous

oo nevermind. understood

- anonymous

Well.. It converges "for sure" when it's less than 1, and diverges when it's greater than one.
But the test fails when it's equal to 1. So we have to check it separately. Does that make sense?

- anonymous

does x also have to be greater than -1?? y is that?

- anonymous

Yeah. because we had IxI<1, that means x is between -1 and 1.

- anonymous

can we always use the ratio test for this and if we do that isnt x always in between -1 and 1?

- anonymous

I have an exam today, and still have a lot of material to study. So, I should get going now :(.
I am sure sstarica will be here to help.

- anonymous

oh ok then! thanks for all the help!

- anonymous

No, you don't have always to use the ratio test. You can use other tests as well, especially the root test. It just depends on the series itself. The radius of convergence differs from series to another. GOOD LUCK!

- anonymous

http://www.youtube.com/watch?v=01LzAU__J-0

- anonymous

thank u so much! good luck to you too!

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