Hint for prove $\forall n \in \Bbb N \sum _{i=1}^{n} \frac {1}{i!} \le 2 - \frac{1}{2^{n-1}}$? I'm trying to prove the following inequality $\forall n \in \Bbb N$:
$$\sum _{i=1}^{n} \frac {1}{i!} \le 2 - \frac{1}{2^{n-1}}$$
I'm doing it by induction. It's true for $P(1)$.
So now I want to prove it for $P(n+1)$.
$$\sum _{i=1}^{n+1} \frac {1}{i!} \le 2 - \frac{1}{2^{n}}$$
I assume the inductive hypothesis for $P(n)$: $\sum _{i=1}^{n} \frac {1}{i!} \le 2 - \frac{1}{2^{n-1}}$ I'm left with:
$$\sum _{i=1}^{n} \frac {1}{i!} + \frac{1}{(n+1)!} \le 2 - \frac{1}{2^{n}}$$
I'm not assure what step to take next, I should bound it somehow but it's not clear to me how. If anyone could suggest me the next move... thanks.
 A: Hint:
It will be sufficient to show that $$\frac{1}{(n+1)!}\le\frac{1}{2^{n-1}}-\frac{1}{2^n}=\frac{1}{2^n}$$ which follows easily from the fact
$$2^n\le (n+1)!\qquad n\ge 1\text{ integer}\tag{1}$$
Inequality $(1)$ holds since $2\le 1\cdot 2$ and $2< i+1$ for all $1<i\le n$.
A: As Angel's answer notes, an ancillary inequality will be of use here, namely 
$$
2^n\leq(n+1)!,\qquad n\geq1\tag{1}
$$
which may be easily proved as well using induction (here is the meat of the inductive proof for $(1)$):
$$
2^{k+1}=2^k\cdot2\leq(k+1)!\cdot2\leq(k+1)!\cdot(k+2)=(k+2)!.
$$
Hence, 
$$
2^n\leq(n+1)!\Longleftrightarrow\frac{1}{2^n}\geq\frac{1}{(n+1)!}.\tag{2}
$$
Thus, the inductive step of your argument looks like so:
\begin{align}
\sum_{i=1}^{k+1}\frac{1}{i!}&= \sum_{i=1}^k\frac{1}{i!}+\frac{1}{(k+1)!}\tag{by definition}\\[1em]
&\leq 2-\frac{1}{2^{k-1}}+\frac{1}{(k+1)!}\tag{by inductive hypothesis}\\[1em]
&\leq 2-\frac{1}{2^{k-1}}+\frac{1}{2^k}\tag{by $(2)$}\\[1em]
&= 2-\frac{2}{2^k}+\frac{1}{2^k}\tag{rewrite}\\[1em]
&= 2-\frac{1}{2^k}.\tag{simplify}
\end{align}
