If one applies the Ratio Test for the series $$\sum_{n=1}^\infty (-1)^n\;n\;4^n\;x^n$$ gets that $$\left|\frac{x_{n+1}}{x_n}\right|=4|x|\frac{n+1}{n}\to 4|x|.$$ Then, the above limit is less than $1$ iff $$|x|<1/4.$$ One can also check that at $x=\pm\frac{1}{4}$ the series diverges, and therefore the interval of convergence is $(-1/4,1/4)$.
If instead one follows the definition of radius of convergence, since the sequence $$\lbrace\left|(-1)^n\;n\;4^n\right|^{\frac{1}{n}}\rbrace$$ is bounded, then one computes $$\lim\sup\left|(-1)^n\;n\;4^n\right|^{\frac{1}{n}}=\lim\sup 4\;n^{\frac{1}{n}}=4\;e^{\frac{1}{e}},$$ since $\sup n^{\frac{1}{n}}=e^{\frac{1}{e}}$. Hence, the radius of convergence of the series is $$R=\frac{1}{4\;e^{\frac{1}{e}}}.$$ This is contradictory since $\frac{1}{4\;e^{\frac{1}{e}}}<\frac{1}{4}$.
Can anyone spot where I'm making a mistake or being sloppy?