# Show that $(1+\frac{1}{n})^n=\sum_{k=0}^{n}\frac{1}{k!}\Rightarrow \lim_{n\rightarrow\infty}(1+\frac{1}{n})^n=\sum_{k=0}^{\infty}\frac{1}{k!}=:e$

I write out the left term expression and get $$\sum_{k=0}^{n}\binom{n}{k}\bigg(\frac{1}{n}\bigg)^k$$ If I could Show that the k-th term of both sequences is equal I would be done.
I.e what I want to show is

$$\binom{n}{k}\bigg(\frac{1}{n}\bigg)^k=\frac{n\cdot\ldots\big((n-k)+1\big)}{k!}\bigg(\frac{1}{n}\bigg)^k\overset{!}{=}\frac{1}{k!}$$

The approach must be wrong, because the above statement would be equivalent to say that

$$\frac{n\cdot\ldots\big(n-(k-1)\big)}{n^k}=1$$

which is not true.

How can I solve the Problem? I also want to ask if I could Show the left side of the implication in the title, why is the Right side true?

• You might cannot prove that $(1+1/n)^n = \sum_0^n 1/k!$ because such equation cannot hold, if I have remembered correctly. – xbh Jan 23 at 15:36
• It's not true : $(1+\frac{1}{2})^2=\frac{9}{4}$ while $1+\frac{1}{1!}+\frac{1}{2!}=2,5$. – Ayoub Jan 23 at 15:37
• Since $p\implies q$ is true when $p$ is false, you may want to change the title of this post if what you really want to show is $q$, namely the identity about $e$. – user587192 Jan 29 at 18:19

The condition $$a_n=b_n$$ for all $$n$$ is sufficient but not necessary for $$\lim_{n\to \infty}a_n=\lim_{n\to\infty}b_n.$$
Note that $$\lim_{n\to\infty}(1+\frac{1}{n})^n=\sum_{k=0}^\infty\frac{1}{k!}=:e\tag{1}$$ But $$(1+\frac{1}{n})^n=\sum_{k=0}^n\frac{1}{k!}$$ is not even true for $$n=2$$.
Let $$s_n=\sum_{k=0}^n\frac{1}{k!},\quad t_n=(1+\frac{1}{n})^n.$$
Step 1. Use the binomial theorem to show that $$t_n\le s_n$$. (Exercise!) It then follows that $$\limsup_{n\to\infty}t_n\le e.\tag{2}$$
Step 2. If $$n\ge m$$, (applying the binomial theorem again) $$t_n\ge 1+1+\frac{1}{2!}(1-\frac{1}{n})+\cdots+\frac{1}{m!}(1-\frac1n)\cdots(1-\frac{m-1}{n}).$$ It follows that for any fixed $$m$$, $$s_m\le\liminf_{n\to\infty} t_n.$$ Letting $$m\to\infty$$, one gets $$e\le\liminf_{n\to\infty} t_n.\tag{3}$$ You get the desired result by combining (2) and (3).