Answer
$0.15833$
Work Step by Step
We will use Remainder Alternating series such as: $\text{Error}=|R_n| \leq a_{n+1}$
Here, we have
$|R_n| \leq \dfrac{1}{(2n+1)!} \lt 10^{-3}$
and $ (2n+1)! \gt 10^3$
or, $n =3$
This implies that the value for $n$ is $n=3$ and the remainder is less than $10^{-3}$.
Next, the value of series is: $\sum_{k=1}^{3} \dfrac{(-1)^{k+1}}{(2k+1)!}=\dfrac{1}{3 !}-\dfrac{1}{5!}=0.15833$
