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.


Let's get this question off the list of unanswered questions. This seems correct.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.