Answer
$-0.4081$
Work Step by Step
We will use Remainder Alternating series such as: $\text{Error}=|R_n| \leq a_{n+1}$
Here, we have
$|R_n| \leq \dfrac{n+1}{(n+1)^4+1} \lt 10^{-3}$
and $ \dfrac{(n+1)^4+1}{n+1} \gt 10^3$
or, $(n+1)^3+\dfrac{1}{n+1} \gt 10^3$
or, $n =9$
This implies that the value for $n$ is $n=9$ and the remainder is less than $10^{-3}$.
Next, the value of series is: $\sum_{k=1}^{10} \dfrac{(-1)^k}{k^4+1}=-0.4081$
