Question

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

Answer 1

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

Answer 2

wait

Answer 3

Just give me a minute.

Answer 4

ok.

Answer 5

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

Answer 6

yes

Answer 7

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

Answer 8

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 :)

Answer 9

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

Answer 10

but it's going to be difficult

Answer 11

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

Answer 12

lol, root test is much easier here :)

Answer 13

let:

\[an = x^n\]

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

Answer 14

and :\[0\le |x| < 1\]

so the radius is = 1 in this case, I guess, correct me if I'm wrong please :)

Answer 15

so since :\[0 \le |x| < 1\]

then :

\[-1 < x < 1\] is the interval of convergence ^_^

Answer 16

I took this section yesterday morning during my calculus class lol

Answer 17

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

Answer 18

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

Answer 19

oh >_<! I thought it was to the power of 1/n

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

oh, almost the same LOL =P bleh

Answer 27

Is the limit part clear?

Answer 28

yess

Answer 29

oh congrats on being a champion anwar ^_^

Answer 30

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

Answer 31

Thanks sstarica :)

Answer 32

:) sure

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

oo nevermind. understood

Answer 37

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?

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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.

Answer 42

oh ok then! thanks for all the help!

Answer 43

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!

Answer 44

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

Answer 45

thank u so much! good luck to you too!