# Prove that $\lim_{n \rightarrow \infty} \sum_{k=0}^{n} \frac{1}{k!} = e$

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.

• See here : math.stackexchange.com/questions/99016/… Commented Aug 9, 2019 at 8:41
• Seems correct to me. Commented Aug 9, 2019 at 8:45
• wow, this was perfectly written (and correct) Commented Aug 9, 2019 at 8:47
• Thank you so much for all of your support ^^ Commented Aug 9, 2019 at 8:50