Answer
The value for $n$ is $n=5$ and the remainder is less than $10^{-4}$.
Work Step by Step
We will use Remainder Alternating series such as: $\text{Error}=|R_n| \leq a_{n+1}$
Here, we have
$R_n \leq \dfrac{1}{4} (\dfrac{1}{4^n})(\dfrac{2}{4n+5}+\dfrac{2}{4n+6}+\dfrac{2}{4n+7}) \lt 10^{-4}$
and $ \dfrac{1}{4^n}(\dfrac{2}{4n+5}+\dfrac{2}{4n+6}+\dfrac{2}{4n+7}) \lt 4 \times 10^{-4}$
or, $n \gt 4.6$
This implies that the value for $n$ is $n=5$ and the remainder is less than $10^{-4}$.
