Define $e, e'$ by $$e: =\lim _{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^{n} \quad \text{and} \quad e' := \lim _{n \rightarrow \infty} \sum_{k=0}^{n} \frac{1}{k!}$$
Prove that $e' \in \mathbb R$ and $e' = e$.
Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!
PS: I've not learned about derivative and logarithmic function yet.
My attempt:
To make the presentation easier to follow, I set $$e_n := \left(1+\frac{1}{n}\right)^{n} \quad e'_n := \sum_{k=0}^{n} \frac{1}{k!}$$
Clearly, $e'_n + 1/((n+1)!) = e'_{n+1}$ and thus the sequence $(e'_n)$ is increasing. On the other hand, $e'_n = 2 + \sum_{k=2}^{n} \frac{1}{k!} \le 2 + \sum_{k=1}^{n} \frac{1}{k(k+1)} = 2+(1-\frac{1}{n+1}) = 3 – \frac{1}{n+1} < 3$. So the sequence $(e'_n)$ is bounded from above and thus $e'$ is well defined.
By binomial theorem, $e_n = \sum_{k=0}^n {n \choose k} \frac{1}{n^k}$ where ${n \choose k} \frac{1}{n^{k}}=\frac{1}{k !} \frac{n \cdot(n-1) \cdot \cdots \cdot(n-k+1)}{n \cdot n \cdot \cdots \cdot n} \leq \frac{1}{k !}$. As such, $e_n \le \sum_{k=0}^n \frac{1}{k !} = e'_n$ and so $e \le e'$. Next we prove that $e \ge e'$.
For $n \ge m$, we have
$$\begin{aligned} e_{n}=\left(1+\frac{1}{n}\right)^{n} &=\sum_{k=0}^{n} {n \choose k} \frac{1}{n^{k}} \geq \sum_{k=0}^{m} {n \choose k} \frac{1}{n^{k}} \\ &=1+\sum_{k=1}^{m} \frac{1}{k !} \frac{n (n-1) \cdots (n-k+1)}{n \cdots n} \\ &= 1+\sum_{k=1}^{m} \frac{1}{k !} \left[ 1 \cdot\left(1-\frac{1}{n}\right) \cdots\left(1-\frac{k-1}{n}\right) \right] \end{aligned}$$
For $n \ge m$, I set $x_{n,m} = 1+\sum_{k=1}^{m} \frac{1}{k !} \left [ 1 \cdot\left(1-\frac{1}{n}\right) \cdots\left(1-\frac{k-1}{n}\right) \right]$. Then $e_n \ge x_{n,m}$ and thus $e = \lim _{n \rightarrow \infty} e_n \ge \lim _{n \rightarrow \infty} x_{n,m} = e'_m$. As such, $e \ge e'$. This completes the proof.
Best Answer
Let's get this question off the list of unanswered questions. This seems correct.