Answer
$-0.4597$
Work Step by Step
The given series is $\Sigma_{n=1}^\infty \dfrac{(-1)^n}{(2n)!}$
The sums of the given series are as follows:
$S_1= \dfrac{-1}{2!}=-0.5 \\S_2=\dfrac{(-1)^1}{2(1)!}+\dfrac{(-1)^2}{2(2)!}\approx -0.4583\\S_3=\dfrac{(-1)^1}{2(1)!}+\dfrac{(-1)^2}{2(2)!}+\dfrac{(-1)^3}{2(3)!}\approx -0.4597\\S_4=\dfrac{(-1)^1}{2(1)!}+\dfrac{(-1)^2}{2(2)!}+\dfrac{(-1)^3}{2(3)!}+\dfrac{(-1)^4}{2(4)!}\approx -0.4597$
It has been seen that the sum of the given series $S_3=S_4$ when approximated up to four decimals.
Thus, we know that the sum of the given series is $-0.4597$ (approximated up to four decimals).