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Topic: \(\textbf{Calculus 2}\)
Q: Radius of convergence and interval of convergence for: \(\large \sum_{n=1}^{\infty} n!(2x1)^n\)
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"}
 one year ago
 one year ago
Topic: \(\textbf{Calculus 2}\) Q: Radius of convergence and interval of convergence for: \(\large \sum_{n=1}^{\infty} n!(2x1)^n\) 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"}
 one year ago
 one year ago

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agentx5Best ResponseYou've already chosen the best response.0
\[\lim_{n \rightarrow \infty} \left \frac{a_{n+1}}{a_n} \right\] \[\lim_{n \rightarrow \infty} \left \frac{(n+1)!(2x1)^{n+1}}{n!(2x+1)^n} \right\] \[\lim_{n \rightarrow \infty} \left \frac{\cancel{n!}(n+1)(2x1)^{\cancel{n}+1}}{\cancel{n!}\cancel{(2x+1)^n}} \right = \infty\]
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
(trying to copy over from the related discussion about the trick shown in the last step to get rid of the factorial)
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
@agentx5, it looks right to me, if you are only asking for a double check.
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Interval of convergence I have correct, 0.5 < x < 0.5 by solving the conjunction" 2x1<0
 one year ago

agentx5Best ResponseYou've already chosen the best response.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...
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
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...
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
I made an educated guess on zero after 1/9 and \(\infty\) failed to be correct
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Graphing this on a TI83 shows it variates, it's not constant at zero.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
well what makes sense to me in the example you just posted @agentx5 is that the term (9x1)^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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
man this is bugging me!!
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
if x=1/n, the the sum converges!! but i haven't seen radius of interval in terms of n
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
yes @experimentX, usually they neatly cancel out.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
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
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
that's the only value for which the series converges and it converges to 0 .. there's no interval ... just a point
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
I believe what @experimentX said would work for both answers @agentx5
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Oh bigger picture question to this then: What IS the radius of convergence, really?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
In your previous example. x=1/2 is the only point where the series converges. And that is one individual point, so no radius.
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
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
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
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
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Ah and by the squeeze theorem it's just 1/9. :)
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Thank you for your assistance gentlemen :)
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
well ... the best would be to use Cauchy root test than D'Alembert's ratio test here!! yw .. if it helped!!
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Whoa! Cauchy what?! D'Alembert what?! Never heard of those lol >_<
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
lots of and lots of convergence test http://en.wikipedia.org/wiki/Convergence_tests
 one year ago
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