$c_n = 1 + \frac{1}{1!} + \frac{1}{2!} +...+\frac{1}{n!}$ 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?

We know
$$e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \cdots$$
Fixing a positive integer $n$, let
\begin{align*}
x &= \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \cdots + \frac{1}{n!}\\[4pt]
r &=  \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \frac{1}{(n+3)!} + \cdots\\[4pt]
\end{align*}
Thus, we have
$$e = x + r$$
and we want to show 
$$r < \frac{1}{n{(n!)}}$$
Then
\begin{align*}
r\, &=  \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \frac{1}{(n+3)!} + \cdots\\[4pt]
\implies\; (n+1)r\, &= \frac{n+1}{(n+1)!} + \frac{n+1}{(n+2)!} + \frac{n+1}{(n+3)!} + \cdots\\[4pt]
&< \frac{n+1}{(n+1)!} + \frac{n+2}{(n+2)!} + \frac{n+3}{(n+3)!} + \cdots\\[4pt]
&= \frac{1}{n!} + \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \cdots\\[4pt]
&= \frac{1}{n!} + r\\[16pt]
\text{Then}\;\,(n+1)r\, &< \frac{1}{n!} + r\\[4pt]
\implies\; nr\, &< \frac{1}{n!}\\[4pt]
\implies\;\;\;\, r\, &< \frac{1}{n{(n!)}}\\[4pt]
\end{align*}
as required.
