Question

Find the interval of convergence of the power series (3^n*x^n)/(n*4^n)

From n=1 to infinity

Answer 1

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

Answer 2

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:
\[R = \frac{4}{3}\]

Answer 3

furthermore, your series converges at x=0

Answer 4

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

Answer 5

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

Answer 6

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