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

wait

Just give me a minute.

ok.

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

yes

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

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

but it's going to be difficult

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

lol, root test is much easier here :)

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

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

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

I took this section yesterday morning during my calculus class lol

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

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

oh >_

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

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

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

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

oh, almost the same LOL =P bleh

Is the limit part clear?

yess

oh congrats on being a champion anwar ^_^

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

Thanks sstarica :)

:) sure

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

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

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

oo nevermind. understood

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

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

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

oh ok then! thanks for all the help!

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

thank u so much! good luck to you too!