Calculating infinite sum with Taylor expansion

Question

I know from WolframAlpha that $ \sum _{n=1}^{\infty }\frac{3^n}{n4^n} = \ln(4)$ but I wasn't able to prove it, I believe that the sum is a sort of manipulation on the Taylor series expansion of $\ln(1+x)$ but didn't found what it is, any hint how to continue?

Answer 1

$$\frac {1}{1-x}＝\sum_{k＝0}^{\infty}x^k\;\implies\;\;\int\;\frac {1}{1-x}dx＝\int\;\left({\sum_{k＝0}^{\infty}x^k}\right)dx\;\\~\\\implies\;-\ln(1-x)＝\sum_{k＝0}^{\infty}\frac {x^{k＋1}}{k＋1}\\~\\\implies\;-\ln(1-x)＝\sum_{k＝1}^{\infty}\frac {x^{k}}{k}\\~\\\implies\;-\ln(1-\frac {3}{4})＝\sum_{k＝1}^{\infty}\frac {\left({\frac {3}{4}}\right)^{k}}{k}\\~\\\implies\;\ln(4)＝\sum_{k＝1}^{\infty}\frac {3^k}{k4^k}$$