Answer
$-0.4597$
Work Step by Step
Given: the series $\Sigma_{n=1}^\infty \dfrac{(-1)^n}{(2n)!}$
$S_1=\Sigma_{n=1}^\infty \dfrac{(-1)^n}{(2n)!}=\dfrac{(-1)^1}{(2(1))!}= \dfrac{-1}{2!}=-0.5$
$S_2=\Sigma_{n=1}^\infty \dfrac{(-1)^n}{(2n)!}=\dfrac{(-1)^1}{(2(1))!}+\dfrac{(-1)^2}{(2(2))!}\approx -0.4583$
$S_3=\Sigma_{n=1}^\infty \dfrac{(-1)^n}{(2n)!}=\dfrac{(-1)^1}{(2(1))!}+\dfrac{(-1)^2}{(2(2))!}+\dfrac{(-1)^3}{(2(3))!}\approx -0.4597$
$S_4=\Sigma_{n=1}^\infty \dfrac{(-1)^n}{(2n)!}=\dfrac{(-1)^1}{(2(1))!}+\dfrac{(-1)^2}{(2(2))!}+\dfrac{(-1)^3}{(2(3))!}+\dfrac{(-1)^4}{(2(4))!}\approx -0.4597$
We can see that $S_3=S_4$ when approximated up to four decimals.
Hence, our answer is: $-0.4597$
