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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

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  1. Michele_Laino
    • one year ago
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    we can rewrite your series, as follows: \[\sum {{a_n}{x^n}} \] where: \[{a_n} = \frac{1}{n}{\left( {\frac{3}{4}} \right)^n}\]

  2. Michele_Laino
    • one year ago
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    now, 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: \[R = \frac{4}{3}\]

  3. Michele_Laino
    • one year ago
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    furthermore, your series converges at x=0

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

  5. anonymous
    • one year ago
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    $$\sum_{n=1}^\infty \frac1n\left(\frac34\right)^n x^n=\sum_{n=0}^\infty \frac1{n+1}\left(\frac34\right)^{n+1}x^{n+1}=\sum_{n=0}^\infty\left(\frac34\right)^{n+1}\int x^n\, dx$$ so $$\int\left(\sum_{n=0}^\infty\left(\frac34\right)^{n+1}x^n\right)dx=\int\frac{3/4}{1-x\cdot 3/4}\, dx=-\log(1-x\cdot3/4)$$

  6. anonymous
    • one year ago
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    we 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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