A calculus problem by Kumar Mayank

From Wikipedia , the identity to use is $\sum\limits_{k=0}^{\infty} \dfrac{z^k}{k!} = e^z$

Expressing in summation form, $\begin{array}{rl} \dfrac{1}{2} - \dfrac{1}{6} + \dfrac{1}{24} - \dfrac{1}{120} &= \dfrac{(-1)^2}{1 \cdot 2} + \dfrac{(-1)^3}{1 \cdot 2 \cdot 3} + \dfrac{(-1)^4}{1 \cdot 2 \cdot 3 \cdot 4} + \dfrac{(-1)^5}{1 \cdot 2 \cdot 3 \cdot 5} + \cdots\\ &= \dfrac{{\color{#3D99F6}(-1)}^2}{{\color{#D61F06}2!}} + \dfrac{{\color{#3D99F6}(-1)}^3}{{\color{#D61F06}3!}} + \dfrac{{\color{#3D99F6}(-1)}^4}{{\color{#D61F06}4!}} + \dfrac{{\color{#3D99F6}(-1)}^5}{{\color{#D61F06}5!}} + \cdots\\ &= \sum\limits_{n=2}^{\infty} \dfrac{{\color{#3D99F6}(-1)^n}}{{\color{#D61F06}n!}} \end{array}$ Observe that $\sum\limits_{n=2}^{\infty} \dfrac{(-1)^n}{n!} = \sum\limits_{n=0}^{\infty} \dfrac{(-1)^n}{n!} - {\color{#20A900}\sum\limits_{n=0}^{2} \dfrac{(-1)^n}{n!}}$ Since $\color{#20A900}\sum\limits_{n=0}^{2} \dfrac{(-1)^n}{n!} = 0$ this concludes that $\sum\limits_{n=2}^{\infty} \dfrac{(-1)^n}{n!} = \sum\limits_{n=0}^{\infty} \dfrac{(-1)^n}{n!} = \boxed{e^{-1}}$