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anonymous
 4 years ago
calculate the radius of convergence and the interval of convergence (with examination of the endpoints)
\[\sum_{n=1}^{\infty}\frac{3^{n+1}}{n}.x^{n}\]
a try...
\[\frac{a_{n+1}}{a_{n}}=\]
\[
\Large \frac{\frac{3^{(n+1)+1}}{n+1}}{\frac{3^{n+1}}{n}}=\]
\[\frac{3^{n+2}}{n+1}.\frac{n}{3^{n+1}}=\frac{3}{1}=3\]
is it correct and what i need to do additional?
anonymous
 4 years ago
calculate the radius of convergence and the interval of convergence (with examination of the endpoints) \[\sum_{n=1}^{\infty}\frac{3^{n+1}}{n}.x^{n}\] a try... \[\frac{a_{n+1}}{a_{n}}=\] \[ \Large \frac{\frac{3^{(n+1)+1}}{n+1}}{\frac{3^{n+1}}{n}}=\] \[\frac{3^{n+2}}{n+1}.\frac{n}{3^{n+1}}=\frac{3}{1}=3\] is it correct and what i need to do additional?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[R=3 \quad 3\le R \le 3\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that means my solution is nearly correct right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry I made a mistake, first find the Limit, in your case L=3 then Radius is R=1/L

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0why we get Radius 1/L ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm ok thank you cinar
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