Hello everyone

 

I've seen 2 definitions for [math]e[/math] (one on the internet, one I have proven myself):

 

[math]e=\lim_{n\to\infty}{\left[1+\frac{1}{n}\right]^n}[/math]

 

and

 

[math]e=\lim_{n\to 0}{\left[n+1\right]^{\frac{1}{n}}}[/math]

 

Now, is there a proof that these two are equal? (I'd first like to know if I can change the first expression to something in the form of [math]n\to 0[/math].)

 

Thanks!

 

Function

Edited November 13, 201312 yr by Function

According to wiki they are provable equivalent

http://en.wikipedia.org/wiki/E_%28mathematical_constant%29#Alternative_characterizations

 

Although I haven't looked at their proof

Maybe I'm nitpicking but surely both these definitions are flawed?

 

e is the sum of an infinite series, which we can prove by Tannery's theorem converges to the first limit.

 

With the second limit what is the first n in your series ?

Hello everyone

 

I've seen 2 definitions for [math]e[/math] (one on the internet, one I have proven myself):

 

[math]e=\lim_{n\to\infty}{\left[1+\frac{1}{n}\right]^n}[/math]

 

and

 

 

 

Now, is there a proof that these two are equal? (I'd first like to know if I can change the first expression to something in the form of [math]n\to 0[/math].)

 

Thanks!

 

Function

You may want to restrict the second one to the right-handed limit, i.e.

[math]e=\lim_{n\to 0^+}{\left[n+1\right]^{\frac{1}{n}}}[/math]

 

Then, we can easily see that the two are equivalent by looking at the first limit and letting [math]n = \frac{1}{n'}[/math].

You may want to restrict the second one to the right-handed limit, i.e.

[math]e=\lim_{n\to 0^+}{\left[n+1\right]^{\frac{1}{n}}}[/math]

 

Then, we can easily see that the two are equivalent by looking at the first limit and letting [math]n = \frac{1}{n'}[/math].

 

Thanks