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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?
 one year ago
 one year 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?
 one year ago
 one year ago

This Question is Closed

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

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

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

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

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