Using $ \sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}=\frac1e$ evaluate first $3$ decimal digits of $1/e$. 
Using the series $\displaystyle \sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}=\frac{1}{e}$, evaluate the first $3$ decimal  digits of $1/e$.

Attempt. In alternating series $\displaystyle \sum_{k=0}^{\infty}(-1)^{k+1}\alpha_n$, where $\alpha_n \searrow 0$, if $\alpha$ is the sum of the series then $$|s_n-\alpha|\leq \alpha_{n+1}.$$ So, in our case  we need to find $n$
such that $|s_n-1/e|<0.001$, where $\displaystyle s_n=\sum_{k=0}^{n-1}\frac{(-1)^{k}}{k!}$ and it is enough to find $n$ such that $\dfrac{1}{n!}<0.001$, so $n\geq 7$. Therefore:
$$s_7=\sum_{k=0}^{6}\frac{(-1)^{k}}{k!}=0.36805\ldots$$
so I would expect $\dfrac{1}{e}=0.368\ldots$. But:
$\dfrac{1}{e}=0.36787944\ldots$.
Where am I missing something?
Thanks in advance.

Note
$$\bigg|\sum_{k=0}^{n}\frac{(-1)^{k}}{k!}-\frac{1}{e}\bigg|=\bigg|\sum_{k=n+1}^{\infty}\frac{(-1)^{k}}{k!}\bigg|\le\sum_{k=n+1}^\infty\frac{1}{3^{k}}\le\frac{1}{2\cdot3^{n}}.$$
Let $\frac{1}{2\cdot3^{n}}<0.001$ and then $n>\frac{\ln500}{\ln 3}\approx5.65678$. Now set $n=6$ and then
$$ \bigg|\sum_{k=0}^{6}\frac{(-1)^{k}}{k!}-\frac{1}{e}\bigg|<0.001. $$
In fact, it is easy to see
$$ \bigg|\sum_{k=0}^{5}\frac{(-1)^{k}}{k!}-\frac{1}{e}\bigg|=0.001212774505,\bigg|\sum_{k=0}^{6}\frac{(-1)^{k}}{k!}-\frac{1}{e}\bigg|=0.0001761143841<0.001. $$
So $n=6$ is the small number such that $|s_n-\frac1e|<0.001$.
