## 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

1. Michele_Laino

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

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

furthermore, your series converges at x=0

4. Michele_Laino

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

$$\sum_{n=1}^\infty \frac1n\left(\frac34\right)^n x^n=\sum_{n=0}^\infty \frac1{n+1}\left(\frac34\right)^{n+1}x^{n+1}=\sum_{n=0}^\infty\left(\frac34\right)^{n+1}\int x^n\, dx$$ 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

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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