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Discussion in 'Mathematics' started by Polecat, Mar 6, 2012.
Drat, DM got there first.
Note the ratio test shows it converges for |x| 1, but says nothing about what happens when |x| = 1.
It seems that to get <
here you need to type
OK, DF, so what do you think is the interval of convergence, as distinct from the radius of convergence?
Never seen "interval of convergence" used before. To be honest my "off the top of the head" answer to this is "don't know, although it's obvious for some choices of n". In practical terms, the behvaiour of the series for |x| less than 1 is so much better than for |x| = 1 (i.e. you can integrate, differentiate and take limits freely) that you want to avoid the |x| = 1 case when possible.
Wiki has the (very specific) convergence details at the radius of convergence:
Gosh, I didn't think I'd have to switch into teaching mode.
The radius of convergence is R where R>0 is such that the series converges for |x| < R.
Now it sometimes happens that the series converges for x = R and/or x = - R, or neither.
Consider the series sigma(x^n/n, n= 1..). Its radius of convergence is R = 1. It converges for x = -1,but not x = 1. Its interval of convergence is [-1, 1).
Consider sigma(x^n/n^2,n=1..). Its interval of convergence is [-1, 1].
Consider sigma(x^n, n= 0..). Its interval of convergence is (-1, 1)
I'm not sure that you did. I know perfectly well what the radius of convergence is, and although I'd not seen "interval of convergence" before, it was obvious what you meant from context. And of course for many series it's obvious what the endpoint behaviour is with little or no calculation.
My point was that for the particular case of the (infinite) binomial series, the endpoint conditions are tricky enough, and are so rarely useful in any practical sense, that I don't personally see any point in remembering them. In the rare cases where you need to know, it's easy enough to work it out by hand.