- anonymous

Power series problem.
(x-2)^n/(n^n)
Find the radius and interval of convergence.

- jamiebookeater

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- aykayyy

i actually just learned this today lol..... try the ratio test!

- anonymous

I did, but am not sure if I am doing ti correctly

- anonymous

I would do the nth root test?

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## More answers

- aykayyy

\[\lim_{n \rightarrow \infty} \left| \frac{ a _{n+1} }{ a _{n} } \right|\]

- anonymous

|dw:1375316023379:dw|

- aykayyy

oh yes @sarahusher is probably right! i didnt see the ^n....
the root test would work better

- anonymous

ok ill try that

- anonymous

ok i did the root test and got (x-2)/n

- aykayyy

then take the limit of that

- anonymous

Yep!

- anonymous

i pull the x-2 out, and the limit of (1/n) would be 0

- anonymous

exactly!

- anonymous

but then i multiply the 0 by the (x-2) right?

- anonymous

which would give me 0 overall?

- aykayyy

yup

- anonymous

As limit < 1, the series will converge for every 'x'

- anonymous

Im not sure how they are getting the interval of convergence which they are saying is from negative infinity to positive infinity

- anonymous

Okay, I'll explain:
So when finding the radius of convergence:
We know that as the limit=0<1, the series is convergent for every 'x'
So for any x for any 'x' you get (ie from -infinity to +infinity) the series will converge
Using the limit that you have, you get to ROC = infinity
which directly gives you the interval as -infinity

- anonymous

Does that make sense?

- anonymous

Everything else makes sense, but the ROC
(x-2) Lim as 'n'-------> infinity of (1/n) = 0
0*(x-2)= 0
Im just not sure how the ROC is infiinty

- anonymous

'The radius of convergence' r is a 'nonnegative real number' or '∞' such that the series converges if [x+L] < r , here L is your limit
So our limit is 0
so we want an 'r' such that all of our 'x'

- anonymous

|dw:1375317501046:dw|
Think of it like a disc, no matter where your value of 'X' lies in the disc, for this example the series will always converge

- anonymous

ok that makes more sense now...thanks sara :)

- anonymous

Okay, If anything isn't clear let me know :)

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