Answer
$n \geq 999$
Work Step by Step
We have $S_n=a_1-a_2+......+(-1)^{n+1}a_n$
Now, $|S-S_n| \leq |a_{n+1}|$
So, $|a_{n+1} | \lt 0.001 =|\dfrac{n+1}{(n+1)^2+1}| \lt 0.001$
or, $(n+1)^2+1 \gt 1000(n+1)$
or, $n \geq 999$
Thus, we need $n \geq 999$ terms to estimate the sum of the entire series when an error is less than $0.001$.