find the precise range of of the following sum I need to find the precise range of the following sum to infinity: 
$$\sum_{n=1}^\infty\frac{x^n}{4^nn}$$
Can someone post how to do it because I have no idea, I'm also given that it converges...
 A: $$f(x) = \sum_{n=1}^{\infty} \dfrac{(x/4)^n}n$$
Consider $g(x) = \displaystyle \sum_{n=0}^{\infty} (x/4)^n$. This converges for $\vert x \vert < 4$ and it converges to $$\dfrac1{1-x/4} = \dfrac{4}{4-x}$$
Note that $$\dfrac{df}{dx} = \sum_{n=1}^{\infty} \dfrac{nx^{n-1}}{n4^n} = \dfrac14 \cdot \sum_{n=0}^{\infty} (x/4)^n = \dfrac14 g(x) = \dfrac1{4-x}$$
(You may want to read why term by term differentiation or integration of a power series within its domain of convergence is valid)
Hence, $$f(x) = \int_0^x \dfrac{dt}{4-t} + f(0)$$ And we know that $f(0) = 0$. Hence, $$f(x) = \int_0^x \dfrac{dt}{4-t} = - \log(4-t) \vert_{0}^x = \log 4 - \log(4-x)$$
Hence, $\displaystyle f(x) = \sum_{n=1}^{\infty} \dfrac{(x/4)^n}n$ converges for $\vert x \vert < 4$ and converges to $\log 4 - \log(4-x)$
A: Let $$f(x)=\sum_{n\ge 1}\frac{x^n}{4^nn}=\sum_{n\ge 1}\frac1n\left(\frac{x}4\right)^n\;,$$ which is easily seen to converge for $-4\le x<4$. Then
$$f\,'(x)=\sum_{n\ge 1}\frac{x^{n-1}}{4^n}=\frac14\sum_{n\ge 0}\left(\frac{x}4\right)^n=\frac14\cdot\frac1{1-x/4}=\frac1{4-x}\;.$$
Now you know exactly what $f\,'(x)$ is, and you know that $f(0)=0$; can you use this information to find a nice closed expression for $f(x)$ and then use that to answer the question?
