We can find the number of terms needed to fall within that accuracy. Taking the absolute difference of the [tex]n[/tex]th and [tex](n+1)[/tex]th partial sums, you have
[tex]\displaystyle\left|\sum_{k=1}^{n+1}\frac{(-1)^k}{(2k+1)!}-\sum_{k=1}^n\frac{(-1)^k}{(2k+1)!}\right|=\frac1{(2n+3)!}<10^{-4}=\dfrac1{10000}[/tex]
or equivalently,
[tex](2n+3)!>10000[/tex]
It's easy enough to check that [tex](2n+2)![/tex] exceeds 10000 for all [tex]n\ge3[/tex], since [tex](2\times3+2)!=8!=362880[/tex]. This means the sum can be approximated to within [tex]10^{-4}[/tex] with just the first three terms.
You have
[tex]\displaystyle\sum_{k=1}^3\frac{(-1)^k}{(2k+1)!}=-\frac1{3!}+\frac1{5!}-\frac1{7!}=-\frac1{7!}\left(\frac{7!}{3!}-\frac{7!}{5!}+1\right)=-\frac{799}{5040}\approx-0.158532[/tex]
The actual value of the sum is [tex]\sin(1)-1[/tex], which comes from the fact that
[tex]\sin x=\displaystyle\sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)!}\implies \sin x-x=\sum_{k=1}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)!}[/tex]
and plugging in [tex]x=1[/tex]. The actual value is approximately [tex]-0.158529[/tex] which verifies that the approximation is within 4 decimal places of the infinite series.
