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anonymous
 one year ago
Find the interval of convergence of the power series (3^n*x^n)/(n*4^n)
From n=1 to infinity
anonymous
 one year ago
Find the interval of convergence of the power series (3^n*x^n)/(n*4^n) From n=1 to infinity

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we can rewrite your series, as follows: \[\sum {{a_n}{x^n}} \] where: \[{a_n} = \frac{1}{n}{\left( {\frac{3}{4}} \right)^n}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, I apply this formula: \[\begin{gathered} R = \frac{1}{L} \hfill \\ \hfill \\ L = \mathop {\max \lim }\limits_{n \to \infty } \sqrt[n]{{\left {{a_n}} \right}} = \mathop {\max \lim }\limits_{n \to \infty } \sqrt[n]{{\frac{1}{n}{{\left( {\frac{3}{4}} \right)}^n}}} = \frac{3}{4} \hfill \\ \end{gathered} \] so, we have: \[R = \frac{4}{3}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1furthermore, your series converges at x=0

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1and for x=4/3, your series becomes the geometrical series: \[\sum {\frac{1}{n}} \] which doesn't converges. So your series converges in: \[\left( {  \frac{4}{3},\frac{4}{3}} \right)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$\sum_{n=1}^\infty \frac1n\left(\frac34\right)^n x^n=\sum_{n=0}^\infty \frac1{n+1}\left(\frac34\right)^{n+1}x^{n+1}=\sum_{n=0}^\infty\left(\frac34\right)^{n+1}\int x^n\, dx$$ so $$\int\left(\sum_{n=0}^\infty\left(\frac34\right)^{n+1}x^n\right)dx=\int\frac{3/4}{1x\cdot 3/4}\, dx=\log(1x\cdot3/4)$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we didn't actually need to do the integral or any of that, but the point is that if \(f\) is analytic at \(x_0\) then its derivative \(f'\) is also analytic at \(x_0\) with the same radius of convergence; since our series is the derivative of a geometric one, and the geometric series has a radius of convergence \(x<4/3\), our series also has the same radius of convergence
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