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


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


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?

  • 1
    $\begingroup$ You might cannot prove that $(1+1/n)^n = \sum_0^n 1/k!$ because such equation cannot hold, if I have remembered correctly. $\endgroup$ – xbh Jan 23 at 15:36
  • $\begingroup$ It's not true : $(1+\frac{1}{2})^2=\frac{9}{4}$ while $1+\frac{1}{1!}+\frac{1}{2!}=2,5$. $\endgroup$ – Ayoub Jan 23 at 15:37
  • $\begingroup$ 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 $e$. $\endgroup$ – 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 $n=2$.

Here is a sketch for one way to show (1).

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).


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