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Requiem
Power series problem. (x-2)^n/(n^n) Find the radius and interval of convergence.
i actually just learned this today lol..... try the ratio test!
I did, but am not sure if I am doing ti correctly
I would do the nth root test?
\[\lim_{n \rightarrow \infty} \left| \frac{ a _{n+1} }{ a _{n} } \right|\]
oh yes @sarahusher is probably right! i didnt see the ^n.... the root test would work better
ok i did the root test and got (x-2)/n
then take the limit of that
i pull the x-2 out, and the limit of (1/n) would be 0
but then i multiply the 0 by the (x-2) right?
which would give me 0 overall?
As limit < 1, the series will converge for every 'x'
Im not sure how they are getting the interval of convergence which they are saying is from negative infinity to positive infinity
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
Does that make sense?
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
'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
|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
ok that makes more sense now...thanks sara :)
Okay, If anything isn't clear let me know :)