The sum of the series $$\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \ldots$$ upto infinity is

The given series is

$$S=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\cdots$$

Write the general term: for $$k=0,1,2,\ldots$$ the $$k^{\text{th}}$$ term is $$\dfrac{(-1)^k}{(k+2)!}$$, so

$$S=\sum_{k=0}^{\infty}\frac{(-1)^k}{(k+2)!}\;.$$

Recall the Maclaurin expansion of the exponential function:

$$e^{x}=\sum_{n=0}^{\infty}\frac{x^{\,n}}{n!}\;.$$

Put $$x=-1$$ to get

$$e^{-1}=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}=1-1+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\cdots\;.$$

Observe that the first two terms $$1-1$$ cancel out to zero, and the remaining series is exactly $$S$$:

$$e^{-1}=0+\Bigl(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\cdots\Bigr)=S\;.$$

Hence

$$S=e^{-1}\;.$$

Option B which is: $$e^{-1}$$