Why is$ (1+\frac{1}{n})^n=e$ when n goes to infinity? [duplicate]
This question already has an answer here:
How come such different methods result in the same number, $e$? 2 answers
Why is $\lim\limits_{n\to\infty}(1+\frac1n)^n=e$?
I think it involves $\sum\limits_{n=0}^\infty\frac1{k!}=e$ but not sure
how to get from one to the other.
No comments:
Post a Comment