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Topic: \(\textbf{Calculus 2}\) Q: Radius of convergence and interval of convergence for: \(\large \sum_{n=1}^{\infty} n!(2x-1)^n\) Ratio Test for Divergence \(\lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right|\) = L if ( L < 1 ) then {convergent} if( L > 1 ) then {"divergent"} else {"inconclusive"}

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\[\lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right|\] \[\lim_{n \rightarrow \infty} \left| \frac{(n+1)!(2x-1)^{n+1}}{n!(2x+1)^n} \right|\] \[\lim_{n \rightarrow \infty} \left| \frac{\cancel{n!}(n+1)(2x-1)^{\cancel{n}+1}}{\cancel{n!}\cancel{(2x+1)^n}} \right| = \infty\]
(trying to copy over from the related discussion about the trick shown in the last step to get rid of the factorial)
@agentx5, it looks right to me, if you are only asking for a double check.

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Other answers:

Interval of convergence I have correct, -0.5 < x < 0.5 by solving the conjunction" |2x-1|<0
But the radius? Why is it zero? (correct answer) I would have said the radius is either the same as the interval or all real numbers...
See what I mean @Spacelimbus ? This is the second time I've seen this type of problem, so there's some concept here about radius of convergence I'm not understanding I guess...
I made an educated guess on zero after 1/9 and \(\infty\) failed to be correct
Graphing this on a TI-83 shows it variates, it's not constant at zero.
well what makes sense to me in the example you just posted @agentx5 is that the term (9x-1)^n approaches it's limit faster then n! , but then I would have said that this is only the case if it it smaller than 1.
man this is bugging me!!
if x=1/n, the the sum converges!! but i haven't seen radius of interval in terms of n
yes @experimentX, usually they neatly cancel out.
yeah they do ... still, for any numerical value of x, the series clearly diverges no matther how small ... but for variable value 1/n ... this series converges. I haven't seen radius of convergence expressed interms of n either ... and can't show from from ratio test either
1/9
that's the only value for which the series converges and it converges to 0 .. there's no interval ... just a point
I believe what @experimentX said would work for both answers @agentx5
Oh bigger picture question to this then: What IS the radius of convergence, really?
In your previous example. x=1/2 is the only point where the series converges. And that is one individual point, so no radius.
Like in a practical or visual sense so I can understand it, what is it? Interval is the range of inputs in which it would converge
it's a point
A radius of convergence would mean, any value which is inside that radius (think of a circle) would make the sum converge, but in the examples you have posted, there is only one point that makes that happen, a point has no radius.
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radius of convergence is an interval for which the the series converges, for eg sum, 2^n x^n if |x| < 1/2 .. the series converges -1/2 < x < 1/2 in out case radius of convergence is 1/9 <= x <= 1/9
Ah and by the squeeze theorem it's just 1/9. :-)
Thank you for your assistance gentlemen :-)
well ... the best would be to use Cauchy root test than D'Alembert's ratio test here!! yw .. if it helped!!
Whoa! Cauchy what?! D'Alembert what?! Never heard of those lol >_<
I'm impressed!
lots of and lots of convergence test http://en.wikipedia.org/wiki/Convergence_tests

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