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Power series problem.
(x2)^n/(n^n)
Find the radius and interval of convergence.
 8 months ago
 8 months ago
Power series problem. (x2)^n/(n^n) Find the radius and interval of convergence.
 8 months ago
 8 months ago

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aykayyyBest ResponseYou've already chosen the best response.0
i actually just learned this today lol..... try the ratio test!
 8 months ago

RequiemBest ResponseYou've already chosen the best response.0
I did, but am not sure if I am doing ti correctly
 8 months ago

sarahusherBest ResponseYou've already chosen the best response.3
I would do the nth root test?
 8 months ago

aykayyyBest ResponseYou've already chosen the best response.0
\[\lim_{n \rightarrow \infty} \left \frac{ a _{n+1} }{ a _{n} } \right\]
 8 months ago

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

RequiemBest ResponseYou've already chosen the best response.0
ok i did the root test and got (x2)/n
 8 months ago

aykayyyBest ResponseYou've already chosen the best response.0
then take the limit of that
 8 months ago

RequiemBest ResponseYou've already chosen the best response.0
i pull the x2 out, and the limit of (1/n) would be 0
 8 months ago

RequiemBest ResponseYou've already chosen the best response.0
but then i multiply the 0 by the (x2) right?
 8 months ago

RequiemBest ResponseYou've already chosen the best response.0
which would give me 0 overall?
 8 months ago

sarahusherBest ResponseYou've already chosen the best response.3
As limit < 1, the series will converge for every 'x'
 8 months ago

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

sarahusherBest ResponseYou've already chosen the best response.3
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<x<infinity
 8 months ago

sarahusherBest ResponseYou've already chosen the best response.3
Does that make sense?
 8 months ago

RequiemBest ResponseYou've already chosen the best response.0
Everything 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
 8 months ago

sarahusherBest ResponseYou've already chosen the best response.3
'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
 8 months ago

sarahusherBest ResponseYou've already chosen the best response.3
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
 8 months ago

RequiemBest ResponseYou've already chosen the best response.0
ok that makes more sense now...thanks sara :)
 8 months ago

sarahusherBest ResponseYou've already chosen the best response.3
Okay, If anything isn't clear let me know :)
 8 months ago
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