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- Thread starter okevino
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quasar987

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In any case, you could try to break the thing with l'Hospital.

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HallsofIvy

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[tex]\frac{(n+3)ln(n+3)}{(n+2)ln(n+2)}[/tex]

but once you had that, your "radius of convergence"

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!

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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"isthe 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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