Find an equivalent to $\sum_{k=n}^\infty \frac{1}{k!}$ I would like to find an equivalent of $\sum_{k=n}^\infty \frac{1}{k!}$. Can I do as follow ? (since I always have doubt with those $o$ and $O$, I would like your opinion.

$$n!\sum_{k=n}^\infty \frac{1}{k!}=1+\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\cdots$$
$$=1+\frac{1}{n+1}+\mathcal O\left(\sum_{k=n+1}^\infty \frac{1}{k^2}\right)=1+o(1).$$
and thus the claim follow.

Q1) Is it correct ?

Q2)

I have that $\mathcal O\left(\sum_{k=n+1}^\infty \frac{1}{k^2}\right)$ because $$\frac{1}{(n+1)(n+2)}\leq \frac{1}{(n+1)^2},\quad \frac{1}{(n+1)(n+2)(n+3)}\leq \frac{1}{(n+2)^2},\quad \frac{1}{(n+1)(n+2)(n+3)(n+4)}\leq\frac{1}{(n+3)^2} \cdots$$

and so $\frac{1}{(n+1)(n+2)\cdots(n+k)}\leq \frac{1}{(n+k)^2}$. But suppose we could only prove that $$\frac{1}{(n+1)(n+2)\cdots(n+k)}\leq \frac{C_k}{(n+k)^2},$$
i.e. that the constant depend on $k$ and we can't have a uniform bound. Could I conclude that $$ \frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\cdots=\mathcal O\left(\sum_{k=n+1}^\infty \frac{1}{k^2}\right) \ \ ?$$
See this post: Calculate $\lim_{n \rightarrow \infty}$ ($n!e-[n!e]$)?

It follows that the integer part of $(n-1)!e$ is $(n-1)!\sum\limits_{k=1}^{n-1} \frac{1}{k!}$, so the fractional part  of $(n-1)!e$ is $(n-1)!\sum\limits_{k=n}^{\infty} \frac{1}{k!}$. 
Hence, an equivalent expression is $\frac{\{(n-1)!e\}}{(n-1)!}$.
See this post: Calculate $\lim_{n \rightarrow \infty}$ ($n!e-[n!e]$)?

It follows that the integer part of $(n-1)!e$ is $(n-1)!\sum\limits_{k=1}^{n-1} \frac{1}{k!}$, so the fractional part  of $(n-1)!e$ is $(n-1)!\sum\limits_{k=n}^{\infty} \frac{1}{k!}$. 
Hence, an equivalent expression is $\frac{\{(n-1)!e\}}{(n-1)!}$.