$c_n = 1 + \frac{1}{1!} + \frac{1}{2!} +...+\frac{1}{n!}$ 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?
 A: 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
$$ \left(\frac1{n+1} + \frac1{(n+1)(n+2)} + \cdots \right) < \left( \frac1{n+1} + \frac1{(n+1)^2} + \frac1{(n+1)^3} \cdots \right) $$
$$ = \frac{\frac1{n+1}}{1-\frac1{n+1}} = \frac1{n+1-1} = \frac1n.$$
A: 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!
A: Hint: show that the sequence $c_n$ is strictly increasing and converges to $e$, then show that the sequence $c_n+\frac{1}{n!\cdot n}$ is strictly decreasing and also converges to $e$.
A: 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.
