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agentx5
Group Title
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"}
 2 years ago
 2 years ago
agentx5 Group Title
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"}
 2 years ago
 2 years ago

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agentx5 Group TitleBest ResponseYou've already chosen the best response.0
\[\lim_{n \rightarrow \infty} \left \frac{a_{n+1}}{a_n} \right\] \[\lim_{n \rightarrow \infty} \left \frac{(n+1)!(2x1)^{n+1}}{n!(2x+1)^n} \right\] \[\lim_{n \rightarrow \infty} \left \frac{\cancel{n!}(n+1)(2x1)^{\cancel{n}+1}}{\cancel{n!}\cancel{(2x+1)^n}} \right = \infty\]
 2 years ago

agentx5 Group TitleBest 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)
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
@agentx5, it looks right to me, if you are only asking for a double check.
 2 years ago

agentx5 Group TitleBest 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
 2 years ago

agentx5 Group TitleBest 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...
 2 years ago

agentx5 Group TitleBest 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...
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
I made an educated guess on zero after 1/9 and \(\infty\) failed to be correct
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
Graphing this on a TI83 shows it variates, it's not constant at zero.
 2 years ago

Spacelimbus Group TitleBest 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.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
man this is bugging me!!
 2 years ago

experimentX Group TitleBest 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
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
yes @experimentX, usually they neatly cancel out.
 2 years ago

experimentX Group TitleBest 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
 2 years ago

experimentX Group TitleBest 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
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
I believe what @experimentX said would work for both answers @agentx5
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
Oh bigger picture question to this then: What IS the radius of convergence, really?
 2 years ago

Spacelimbus Group TitleBest 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.
 2 years ago

agentx5 Group TitleBest 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
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
it's a point
 2 years ago

Spacelimbus Group TitleBest 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.
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
dw:1343149575540:dw
 2 years ago

experimentX Group TitleBest 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
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
Ah and by the squeeze theorem it's just 1/9. :)
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
Thank you for your assistance gentlemen :)
 2 years ago

experimentX Group TitleBest 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!!
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
Whoa! Cauchy what?! D'Alembert what?! Never heard of those lol >_<
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
I'm impressed!
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
lots of and lots of convergence test http://en.wikipedia.org/wiki/Convergence_tests
 2 years ago
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