A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
radius and interval of convergence of the series sqrt(n)*x^n as n goes from 1 to infinity?? please help..
anonymous
 5 years ago
radius and interval of convergence of the series sqrt(n)*x^n as n goes from 1 to infinity?? please help..

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the radius = 0, since x stands alone here :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just give me a minute.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you sure it's the sqrt of n and not the power of n?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well.. I am going to use the ratio test first to find the radius of convergence.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you 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
 5 years ago
Best ResponseYou've already chosen the best response.0but since the question doesn't have any sort of exponential function, then "maybe" you can :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but it's going to be difficult

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes I can, and it won't be difficult. And hello! :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, root test is much easier here :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let: \[an = x^n\] so \[an^{\frac{1}{n}} = (x^n)^{\frac{1}{n}} = x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and :\[0\le x < 1\] so the radius is = 1 in this case, I guess, correct me if I'm wrong please :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so since :\[0 \le x < 1\] then : \[1 < x < 1\] is the interval of convergence ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I took this section yesterday morning during my calculus class lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so I might have a couple of mistakes ^_^" maybe

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just so its clear the series is..\[\sqrt{n}*x ^{n}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh >_<! I thought it was to the power of 1/n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\left {a_{n+1} \over a_n} \right=\left {\sqrt{n+1}. x^{n+1} \over \sqrt{n} x^n} \right=\left x \right.\left {\sqrt{n+1} \over \sqrt n} \right\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then find the limit , oh my bad, I completely misunderstood the question >_<

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0proceed with anwar's answer, it seems correct :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh sorry for the confusion..ok yea i got till there..and then?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0continue anwar, you're on the right track :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So 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
 5 years ago
Best ResponseYou've already chosen the best response.0oh, almost the same LOL =P bleh

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is the limit part clear?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh congrats on being a champion anwar ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good. As I said before, the series is convergent when the limit value is <1. In our case, the radius of convergent is: \[R=\left x \right<1 \implies 1<x<1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the radius of convergence = 1 and the interval is the one on top lol ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now you should check the endpoints. check if the series converges at x=1 and x=1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but i thought it had to be less than one to converge..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oo nevermind. understood

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well.. 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
 5 years ago
Best ResponseYou've already chosen the best response.0does x also have to be greater than 1?? y is that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah. because we had IxI<1, that means x is between 1 and 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can we always use the ratio test for this and if we do that isnt x always in between 1 and 1?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok then! thanks for all the help!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, 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
 5 years ago
Best ResponseYou've already chosen the best response.0thank u so much! good luck to you too!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.