$c_n = 1 + \frac{1}{1!} + \frac{1}{2!} +...+\frac{1}{n!}$ so prove $e - c_n \le \frac{1}{n! * n}$ $c_n = 1 + \frac{1}{1!} + \frac{1}{2!} +...+\frac{1}{n!}$, so $e - c_n \le \frac{1}{n! * n}$
I absolutely have no idea how to solve it, could anyone tell me the approach?

The Taylor series for $e^x$ is given by

$$e^x=\sum_{k=0}^\infty\frac{x^k}{k!}\tag 1$$

Using $(1)$ with $x=1$, we have 
$$\begin{align}
e-c_n&=\sum_{k=n+1}^\infty \frac{1}{k!}\\\\
&\le\sum_{k=n+1}^\infty \frac{1}{k!}\left(1+\color{blue}{\underbrace{\frac{1}{k-1}-\frac1k}_{\text{Positive Terms}}}\right)\\\\
&=\sum_{k=n+1}^\infty \frac{1}{k!}\left(\frac{k}{k-1}-\frac1k\right)\\\\
&=\sum_{k=n+1}^\infty\color{red}{\underbrace{\left(\frac{1}{(k-1)(k-1)!}-\frac{1}{k\,k!}\right)}_{\text{Telescoping Terms}}}\\\\
&=\frac{1}{n\,n!}
\end{align}$$
as was to be shown!
