Answer
$-0.030581$
Work Step by Step
We will use Remainder Alternating series such as: $\text{Error}=|R_n| \leq a_{n+1}$
Here, we have
$R_n \leq \dfrac{1}{(2n+3)^5} \lt 10^{-3}$
and $ (2n+3)^3 \gt 10^3$
or, $2n+3 \gt 10$
or, $n \gt 3.5$
This implies that the value for $n$ is $n=4$ and the remainder is less than $10^{-3}$.
Next, the value of series is: $\sum_{k=1}^\infty \dfrac{(-1)^k}{(2k+1)^3}=-\dfrac{1}{3^3}+\dfrac{1}{5^3}-\dfrac{1}{7^3}+\dfrac{1}{9^3}=-0.030581$
