Find the interval of convergence for $\sum\limits_{n=1}^{\infty}\frac{x^n}{n4^n}$. I am working on this problem, but I am not exactly sure about my answer. Can you help me how to do the steps to find the interval of convergence?
My answer is $L = \left| \frac{x}{4} \right| < 1$.

In the ratio test one evaluates
$$
\lim_{n\to\infty} \left| \frac{(n+1)\text{th term}}{n\text{th term}} \right| = \lim_{n\to\infty} \left| \frac{x^{n+1}/((n+1)4^{n+1}}{x^n/(n4^n)} \right| = \frac {|x|} 4.
$$
So the series converges if $|x|/4<1$ and diverges if $|x|/4>1,$ and so it converges if $-4<x<4$ and diverges if $x<-4$ or $x>4.$
That does not say what happens if $|x|/4=1,$ thus if $x=4$ or $x=-4.$ If $x=4$ the series becomes $\sum_{n=1}^\infty 1/n,$ and that diverges to $+\infty.$ If $x=-4$ then the series becomes $\sum_{n=1}^\infty (-1)^n/n,$ and that converges.
Thus the interval of convergence is $[-4,4).$
