28 июн. 2021 г. · How do we know that the limit of (1+1/n)n exists as n increases without bound? · I assume that limn→+∞(1+1n)n=e. I accept this by definition.

12 мая 2013 г. · If you just need to show that the limn→∞(1+1n)n exists, you can just show that it is an increasing sequence which is bounded above.

9 апр. 2017 г. · I would like to show that limn→∞(1+1/n)n=e, using the binomial theorem and the power series expansion of e. So to be clear: I do not want to use ...

2 янв. 2013 г. · I've tried rewriting 1−1n, and operating limn→∞(1−1n)n=1/limn→∞(1+1n)n but I couldn't prove it. Any hint? Thanks in advance. real-analysis ...

25 сент. 2015 г. · Show activity on this post. If I already know that limn→∞an=+∞. Then how can I prove limn→∞(1+1an)an=e. without involving function limit?

25 мар. 2016 г. · Refining Rudin's proof of lim(1+1n)n=lim∑nk=11k!. Let sn=∑n01k!, tn=(1+1n)n. By the binomial theorem, tn=1+1+12!

14 мая 2017 г. · Short answer is you can separate limits and perform one after another. You can't take the limit of 1 + 1/n and when you have gotten that take the limit of (k)n.

22 авг. 2016 г. · If you, with your definition of e have managed to prove that (1+1n)n≤e≤(nn−1)n, then the rest of your solution is correct. – Arthur. Commented ...

24 авг. 2020 г. · I have this messy proof of limn→∞(1+1n)n=e in my notebook. I can't find it anywhere else, but I need it since the professor accepts only this ...

7 окт. 2013 г. · You can obtain the limit of. (1−1n)n. easily from the one you know, (1+1n)n→e, by noting. 1−1n=n−1n=1(nn−1)=1(1+1n−1). Then you can write.