What is the sum of this alternating series: $\sum_{n=1}^\infty \frac{(-1)^n}{4^nn!}$? I need to find the sum of an alternating series correct to 4 decimal places. The series I am working with is: $$\sum_{n=1}^\infty \frac{(-1)^n}{4^nn!}$$ So far I have started by setting up the inequality: $$\frac{1}{4^nn!}<.0001$$Eventually I arrived at $$n=6$$ giving the correct approximation, which is approximately equal to $$\frac{1}{2,949,120}$$ But this is not the answer WolframAlpha gets.

The sum of this series is known, since it is the expansion of  $\;\mathrm e^{-x}-1\;$ for $x=\frac14$.
