ineedhelpnow said:
i don't understand the second one at all. i don't need to find the error though.
You need to find the number of terms $N$ for which the result of the truncated sum is correct to four decimal places. Which requires that the error be less than $0.00005$
As stated in the previous post this occurs when:
$$\frac{1}{3^{N+1}(N+1)!} <0.00005$$
Which holds for any N\ge 4. Hence:
$$S=\sum_{n=1}^{\infty}\frac{-1}{3^{n} n!} \approx \sum_{n=1}^{4} \frac{-1}{3^{n}n!}$$
and the approximation is correct to four decimal places.
Except it turns out that we have to be a bit more careful because of rounding to that fourth decimal place, so let's look at truncation to 4,5 and 6 terms, define:
$$S(N)=\sum_{n=1}^{N}\frac{-1}{3^{n} n!}$$
Then $S(4)=-0.2834362139918$, $S(5)=-0.2834705075446$, $S(6)=-0.2834686023472$, so it looks like we actually need 5 terms in the sum to get it to round correctly to 4 decimal places..