Answer
$-0.9729$
Work Step by Step
We will use Remainder Alternating series such as: $\text{Error}=|R_n| \leq a_{n+1}$
Here, we have
$R_n \leq \dfrac{1}{(n+1)^5} \lt 10^{-3}$
and $ (n+1)^5 \gt 10^3$
or, $n+1 \gt \sqrt[3]{10^3} \approx 2.98$
This implies that the value for $n$ is $n=3$ and the remainder is less than $10^{-3}$.
Next, the value of series is: $\sum_{k=1}^3\dfrac{(-1)^k}{k^5}=-1+\dfrac{1}{2^5}-\dfrac{1}{3^5}=-0.9729$
