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anonymous
 2 years ago
Power series problem.
(x2)^n/(n^n)
Find the radius and interval of convergence.
anonymous
 2 years ago
Power series problem. (x2)^n/(n^n) Find the radius and interval of convergence.

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anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0i actually just learned this today lol..... try the ratio test!

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0I did, but am not sure if I am doing ti correctly

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0I would do the nth root test?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n \rightarrow \infty} \left \frac{ a _{n+1} }{ a _{n} } \right\]

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1375316023379:dw

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0oh yes @sarahusher is probably right! i didnt see the ^n.... the root test would work better

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0ok i did the root test and got (x2)/n

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0then take the limit of that

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0i pull the x2 out, and the limit of (1/n) would be 0

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0but then i multiply the 0 by the (x2) right?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0which would give me 0 overall?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0As limit < 1, the series will converge for every 'x'

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Im not sure how they are getting the interval of convergence which they are saying is from negative infinity to positive infinity

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Okay, 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<x<infinity

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Does that make sense?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Everything else makes sense, but the ROC (x2) Lim as 'n'> infinity of (1/n) = 0 0*(x2)= 0 Im just not sure how the ROC is infiinty

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0'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'<r but 'x' always converges for x<r And we worked out before that x converges for all values (ie for all R  real numbers) So it's not so much a calculation to work this out, as opposed to seeing that no matter what your value the series will converge no matter where it lies along R

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0dw: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
 2 years ago
Best ResponseYou've already chosen the best response.0ok that makes more sense now...thanks sara :)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Okay, If anything isn't clear let me know :)
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