anonymous
  • anonymous
radius and interval of convergence of the series sqrt(n)*x^n as n goes from 1 to infinity?? please help..
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
the radius = 0, since x stands alone here :)
anonymous
  • anonymous
wait
anonymous
  • anonymous
Just give me a minute.

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anonymous
  • anonymous
ok.
anonymous
  • anonymous
are you sure it's the sqrt of n and not the power of n?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Well.. I am going to use the ratio test first to find the radius of convergence.
anonymous
  • 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
  • anonymous
but since the question doesn't have any sort of exponential function, then "maybe" you can :)
anonymous
  • anonymous
but it's going to be difficult
anonymous
  • anonymous
Yes I can, and it won't be difficult. And hello! :)
anonymous
  • anonymous
lol, root test is much easier here :)
anonymous
  • anonymous
let: \[an = x^n\] so \[|an^{\frac{1}{n}} |= (|x^n)^{\frac{1}{n}} |= |x|\]
anonymous
  • anonymous
and :\[0\le |x| < 1\] so the radius is = 1 in this case, I guess, correct me if I'm wrong please :)
anonymous
  • anonymous
so since :\[0 \le |x| < 1\] then : \[-1 < x < 1\] is the interval of convergence ^_^
anonymous
  • anonymous
I took this section yesterday morning during my calculus class lol
anonymous
  • anonymous
so I might have a couple of mistakes ^_^" maybe
anonymous
  • anonymous
just so its clear the series is..\[\sqrt{n}*x ^{n}\]
anonymous
  • anonymous
oh >_
anonymous
  • 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
  • anonymous
then find the limit , oh my bad, I completely misunderstood the question >_<
anonymous
  • anonymous
proceed with anwar's answer, it seems correct :)
anonymous
  • anonymous
oh sorry for the confusion..ok yea i got till there..and then?
anonymous
  • anonymous
continue anwar, you're on the right track :)
anonymous
  • 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
  • anonymous
oh, almost the same LOL =P bleh
anonymous
  • anonymous
Is the limit part clear?
anonymous
  • anonymous
yess
anonymous
  • anonymous
oh congrats on being a champion anwar ^_^
anonymous
  • 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
  • anonymous
Thanks sstarica :)
anonymous
  • anonymous
:) sure
anonymous
  • anonymous
the radius of convergence = 1 and the interval is the one on top lol ^_^
anonymous
  • anonymous
Now you should check the endpoints. check if the series converges at x=1 and x=-1.
anonymous
  • anonymous
but i thought it had to be less than one to converge..
anonymous
  • anonymous
oo nevermind. understood
anonymous
  • 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
  • anonymous
does x also have to be greater than -1?? y is that?
anonymous
  • anonymous
Yeah. because we had IxI<1, that means x is between -1 and 1.
anonymous
  • anonymous
can we always use the ratio test for this and if we do that isnt x always in between -1 and 1?
anonymous
  • 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
  • anonymous
oh ok then! thanks for all the help!
anonymous
  • 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
  • anonymous
http://www.youtube.com/watch?v=01LzAU__J-0
anonymous
  • anonymous
thank u so much! good luck to you too!

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