<s>[INST] $c_n = 1 + \frac{1}{1!} + \frac{1}{2!} +...+\frac{1}{n!}$ 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?
 [/INST] Hint. Using the infinite summation representation of $e$, we have
$$ e - c_n = \sum_{r=0}^\infty \frac1{r!} - \sum_{r=0}^n \frac1{r!} = \sum_{r=n+1}^\infty \frac1{r!} = \frac1{n!}\left(\frac1{n+1} + \frac1{(n+1)(n+2)} + \cdots \right). $$
All that's left to do now is to show that the sum in brackets $()$ is $\le 1/n$.

As the OP pointed out in the comment below, we simply use that $n+r > n+1$ for $r > 1$, and hence
$$ \left(\frac1{n+1} + \frac1{(n+1)(n+2)} + \cdots \right) < \left( \frac1{n+1} + \frac1{(n+1)^2} + \frac1{(n+1)^3} \cdots \right) $$
$$ = \frac{\frac1{n+1}}{1-\frac1{n+1}} = \frac1{n+1-1} = \frac1n.$$
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