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 $L = \left| \frac{x}{4} \right| < 1$.

A quick fomula for power series:
If $a_n=n^{b}\left(\frac{1}{a}\right)^n$ where $a,b\in\mathbb{R}\wedge a\neq0$ we have
$$\sum _{n=1}^{\infty }\:n^{b}\left(\frac{1}{a}\right)^n\left(x-c\right)^n\text{ which converges on }\left\{\begin{array}{l}
(c-a,c+a),a\in\mathbb{R}\wedge b\ge0
\\ [c-a,c+a),a>0\wedge -1\le b<0
\\ (c-a,c+a],a<0\wedge -1\le b<0
\\ [c-a,c+a],a\in\mathbb{R}\wedge b<-1\end{array}\right.$$

$\sum\limits_{n=1}^{\infty}\dfrac{x^n}{n4^n}=\sum\limits_{n=1}^{\infty}n^{-1}\left(\dfrac{1}{4}\right)^n(x-0)^n$ that ,$a=4>0\land b=-1<0$ converges on $[0-4,0+4)$.
