Help in proving $ \left( 1 + \frac{1}{n} \right)^{n} \leq \sum\limits_{k=0}^{n} \frac{1}{k!} < 3 $. I am trying to prove this statement for all $ n \geq 1 $ using induction:
$$
\left( 1 + \frac{1}{n} \right)^{n} \leq \sum_{k=0}^{n} \frac{1}{k!} < 3.
$$
I said:


*

*Base case $ n = 1 $:
$$
  \left( 1 + \frac{1}{1} \right)^{1} \leq \sum_{k=0}^{1} \frac{1}{k!} < 3,
  $$
which is okay.

*Induction step: Suppose that $ \displaystyle \left( 1 + \frac{1}{n}
  \right)^{n} \leq \sum_{k=0}^{n} \frac{1}{k!} < 3 $ for a given $ n \in
  \mathbb{N} $.
Transition from $ n \to n + 1 $:
$$
    \displaystyle \left( 1 + \frac{1}{n + 1} \right)^{n+1}
  = \left( 1 + \frac{1}{n + 1} \right)^{n} \left( 1 + \frac{1}{n + 1} \right)
  = \ldots \text{Help} \ldots
  < 3.
  $$
I need some guidance for proof-writing (-thinking) in orders.
 A: Hint: There is no real need for induction. Use the Binomial Theorem. 
A: For all $ n \in \mathbb{N} $, we have
\begin{align}
      \left( 1 + \frac{1}{n} \right)^{n}
&=    \sum_{k=0}^{n} \binom{n}{k} \left( \frac{1}{n} \right)^{k} \quad (\text{By the Binomial Theorem.}) \\
&=    \sum_{k=0}^{n} \frac{n!}{k!(n - k)!} \cdot \frac{1}{n^{k}} \quad (\text{By the definition of the binomial coefficient.}) \\
&=    \sum_{k=0}^{n} \frac{1}{k!} \cdot \frac{n!}{(n - k)!} \cdot \frac{1}{n^{k}} \\
&=    \sum_{k=0}^{n} \frac{1}{k!} \left( \prod_{i=n-k+1}^{n} i \right) \frac{1}{n^{k}} \quad (\text{By cancellation of terms.}) \\
&\leq \sum_{k=0}^{n} \frac{1}{k!} \left( \prod_{i=n-k+1}^{n} n \right) \frac{1}{n^{k}} \quad (\text{As $ i \leq n $ for all $ i \in \{ n - k + 1,\ldots,n \} $.}) \\
&=    \sum_{k=0}^{n} \frac{1}{k!} \cdot n^{k} \cdot \frac{1}{n^{k}} \\
&=    \sum_{k=0}^{n} \frac{1}{k!} \\
&\leq 1 + \sum_{k=0}^{n-1} \frac{1}{2^{k}} \quad (\text{By comparison of terms.}) \\
&<    1 + \sum_{k=0}^{\infty} \frac{1}{2^{k}} \\
&=    1 + 2 \quad (\text{Sum of a well-known convergent geometric series.}) \\
&=    3. \quad (\text{Voilà!})
\end{align}
A: Your induction hypothesis $(1+\frac{1}{n})^n\leq\sum_{k=0}^{n}\frac{1}{k!}\lt 3$ for a given $n$, not for all $n$.
In the induction step, since $1+\frac1{n+1}>0$, you can multiply both sides of the i.p. by $1+\frac1{n+1}$, to get 
$$\left(1+\frac{1}{n+1}\right)^{n+1}\leq\left(1+\frac{1}{n}\right)^{n}\left(1+\frac{1}{n+1}\right)\overset{\mbox{i.p.}}{\leq}\left(\sum_{k=0}^{n}\frac{1}{k!}\right)\left(1+\frac{1}{n}\right)=\sum_{k=0}^{n}\frac{1}{k!}+\frac1{n+1}\sum_{k=0}^{n}\frac{1}{k!}$$
Can you continue?
A: To show that $\displaystyle\sum_{k=0}^{k=n}\frac{1}{k!} \lt 3$
$\displaystyle\sum_{k=0}^{k=n}\frac{1}{k!}=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{n!}$
$\displaystyle\sum_{k=0}^{k=n}\frac{1}{k!}=1+1+\frac{1}{2!}+\frac{1}{3!}+\dots +\frac{1}{n!}$
$\displaystyle\sum_{k=0}^{k=n}\frac{1}{k!}=1+1+\frac{1}{2}+\frac{1}{2}\left(\frac{1}{3}+\frac{1}{3*4}+\frac{1}{3*4*5}+\dots+\frac{1}{3*4*5*\dots*n}\right)$
$\displaystyle\sum_{k=0}^{k=n}\frac{1}{k!}\lt1+1+1$
