Power series

  • Thread starter okevino
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  • #1
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please take a look at the attachment...

i'm did the |an+1/an|
and then i dont' know how to deal with ln
 

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  • #2
quasar987
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You've got the fraction backward. I.e. your numerator should be your denominator and vice versa.

In any case, you could try to break the thing with l'Hospital.
 
  • #3
HallsofIvy
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I take it you are attempting to find the radius of convergence. Whether you have the fraction upside down or not depends on exactly "where" you are. Certainly to use the "ratio test", you would form
[tex]\frac{(n+3)ln(n+3)}{(n+2)ln(n+2)}[/tex]
but once you had that, your "radius of convergence" is the reciprocal of the limit of that- the limit of the fraction you have.

Obviously the limit of (n+2)/(n+3) (or (n+3)/(n+2)) is 1 so everything depends on the log terms. Treat it as a continuous problem: ln(x+3)/ln(x+2) and use L'Hopital's rule. As it turns out, it really doesn't matter which way you have the fraction!
 
  • #4
quasar987
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Certainly to use the "ratio test", you would form
[tex]\frac{(n+3)ln(n+3)}{(n+2)ln(n+2)}[/tex]
but once you had that, your "radius of convergence" is the reciprocal of the limit of that- the limit of the fraction you have.

Are you sure? in the ratio test, we look at the limit of [tex]|a_{n+1}/a_n|[/tex], and the radius of convergence is the reciprocal of this. Here,

[tex]a_n=\frac{1}{(n+2)\ln(n+2)}[/tex],

so,

[tex]\left|\frac{a_{n+1}}{a_{n}}\right|=\frac{(n+2)\ln(n+2)}{(n+3)\ln(n+3)}[/tex]
 

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