So here is my question. I used the Alternating series to prove that the sum does converge absolutely, so it converges in general.
I then tried to say that $\frac{1}{n^4}<10^{-8}$ and solve for n to find the minimum number of terms. Is that correct or am I doing something wrong?
Hint. A standard estimate of the remainder. One may recall that for a standard alternating series the absolute value of the remainder is less or equal to the absolute value of its first term, giving here $$ \left|R_k:=\sum_{n=k+1}^\infty \frac{(-1)^{n-1}}{n^4}\right|\le \frac1{(k+1)^4}. $$ One may then solve $$ \frac1{(k+1)^4}\le 10^{-8} \implies k=99. $$ Remark. One may prove that $$ \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^4}=\left(1-\frac18\right)\zeta(4)=\frac{7\pi^4}{720}, $$ and that $$ \left|\frac{7\pi^4}{720}-R_{83}\right|>10^{-8},\quad \left|\frac{7\pi^4}{720}-R_{84}\right|\le10^{-8}. $$ The least value is $k=\color{red}{84}.$