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anonymous
 3 years 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"}
anonymous
 3 years 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"}

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

anonymous
 3 years ago
Best 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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@agentx5, it looks right to me, if you are only asking for a double check.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Interval of convergence I have correct, 0.5 < x < 0.5 by solving the conjunction" 2x1<0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But 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...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0See 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...

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Graphing this on a TI83 shows it variates, it's not constant at zero.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well 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.

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes @experimentX, usually they neatly cancel out.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2yeah 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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I believe what @experimentX said would work for both answers @agentx5

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0In your previous example. x=1/2 is the only point where the series converges. And that is one individual point, so no radius.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Like 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0A 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1343149575540:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2radius 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ah and by the squeeze theorem it's just 1/9. :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you for your assistance gentlemen :)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Whoa! Cauchy what?! D'Alembert what?! Never heard of those lol >_<

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