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tunahan
Group Title
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?
 2 years ago
 2 years ago
tunahan Group Title
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?
 2 years ago
 2 years ago

This Question is Closed

cinar Group TitleBest ResponseYou've already chosen the best response.1
\[R=3 \quad 3\le R \le 3\]
 2 years ago

tunahan Group TitleBest ResponseYou've already chosen the best response.0
that means my solution is nearly correct right?
 2 years ago

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

tunahan Group TitleBest ResponseYou've already chosen the best response.0
why we get Radius 1/L ?
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.1
it is definition..
 2 years ago

tunahan Group TitleBest ResponseYou've already chosen the best response.0
hmm ok thank you cinar
 2 years ago
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