Answer
$-0.2691$
Work Step by Step
We will use Remainder Alternating series such as: $\text{Error}=|R_n| \leq a_{n+1}$
Here, we have
$|R_n| \leq \dfrac{n+1}{(n+1)^2+1} \lt 10^{-3}$
and $ \dfrac{(n+1)^2+1}{n+1} \gt 10^3$
or, $n+1+\dfrac{1}{n+1} \gt 10^3$
or, $n =999$
This implies that the value for $n$ is $n=4$ and the remainder is less than $10^{-3}$.
Next, the value of series is: $\sum_{k=1}^{1000} \dfrac{(-1)^k}{k^2+1}=-0.2691$
