- anonymous

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"}

- schrodinger

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- anonymous

\[\lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right|\]
\[\lim_{n \rightarrow \infty} \left| \frac{(n+1)!(2x-1)^{n+1}}{n!(2x+1)^n} \right|\]
\[\lim_{n \rightarrow \infty} \left| \frac{\cancel{n!}(n+1)(2x-1)^{\cancel{n}+1}}{\cancel{n!}\cancel{(2x+1)^n}} \right| = \infty\]

- anonymous

(trying to copy over from the related discussion about the trick shown in the last step to get rid of the factorial)

- anonymous

@agentx5, it looks right to me, if you are only asking for a double check.

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## More answers

- anonymous

Interval of convergence I have correct, -0.5 < x < 0.5 by solving the conjunction"
|2x-1|<0

- anonymous

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...

- anonymous

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...

##### 1 Attachment

- anonymous

I made an educated guess on zero after 1/9 and \(\infty\) failed to be correct

- anonymous

Graphing this on a TI-83 shows it variates, it's not constant at zero.

- anonymous

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.

- experimentX

man this is bugging me!!

- experimentX

if x=1/n, the the sum converges!! but i haven't seen radius of interval in terms of n

- anonymous

yes @experimentX, usually they neatly cancel out.

- experimentX

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

- experimentX

1/9

- experimentX

that's the only value for which the series converges
and it converges to 0 .. there's no interval ... just a point

- anonymous

I believe what @experimentX said would work for both answers @agentx5

- anonymous

Oh bigger picture question to this then: What IS the radius of convergence, really?

- anonymous

In your previous example. x=1/2 is the only point where the series converges. And that is one individual point, so no radius.

- anonymous

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

- experimentX

it's a point

- anonymous

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.

- anonymous

|dw:1343149575540:dw|

- experimentX

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

- anonymous

Ah and by the squeeze theorem it's just 1/9. :-)

- anonymous

Thank you for your assistance gentlemen :-)

- experimentX

well ... the best would be to use Cauchy root test than D'Alembert's ratio test here!!
yw .. if it helped!!

- anonymous

Whoa! Cauchy what?! D'Alembert what?! Never heard of those lol >_<

- anonymous

I'm impressed!

- experimentX

lots of and lots of convergence test
http://en.wikipedia.org/wiki/Convergence_tests

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