How to Prove Euler's Number Converges?

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Discussion Overview

The discussion revolves around methods to demonstrate the convergence of Euler's number \( e \), particularly through the expression \( e = (1 + 1/n)^n \) as \( n \) approaches infinity. Participants explore various proofs and approaches, including the use of the binomial theorem, monotonicity, and series convergence.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests using the binomial formula to express \( (1 + 1/n)^n \) and argues that it is monotonic increasing and bounded above by a convergent series.
  • Another participant proposes using the geometric mean and arithmetic mean inequality to show monotonicity.
  • Some participants note that the convergence of \( \sum 1/n! \) implies the convergence of \( e \), but emphasize the need to first prove the convergence of the original series.
  • Concerns are raised about the clarity of the term "convergent" when referring to numbers, suggesting that it is more appropriate to discuss the convergence of limits or series.
  • There is a challenge regarding the bounding of \( (1 + 1/n)^{n+1} \) and its implications for convergence.

Areas of Agreement / Disagreement

Participants express differing views on the clarity of terminology and the validity of certain proofs. While some methods are acknowledged as valid, no consensus is reached on a single definitive proof or approach.

Contextual Notes

Some participants highlight the need for additional proofs regarding the convergence of the original series and the limits involved, indicating that assumptions and definitions may affect the discussion.

Who May Find This Useful

Readers interested in mathematical proofs, particularly in the context of limits and series convergence, may find this discussion relevant.

jetplan
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Hi All,

How do we go about showing the euler's number e converges ?
Recall that

e =(1+1/n)^n as n ->infinity

Some place prove this by showing the sequence is bounded above by 3 and is monotonic increasing, thus a limit exist.

But I forgot how exactly the proof looks like.


Thanks
J
 
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Here's one way.

Write down (1+1/n)^n using the binomial formula.

You'll see that it equals to

1 + 1 + (1/2!)(1-1/n) + (1/3!) (1-1/n) (1-2/n) + ... + (1/(n-1)!) (1-1/n) (1-2/n) .. (1-(n-1)/n)

From here it's not hard to see that it's monotonic increasing, and it's bounded above by 1 + 1 + 1/2! + 1/3! + ... 1/n!.

In turn, it's not hard to see that the series \sum 1/n! converges, because 1/n! < 2^{-n} for all n>3.
 
Use the fact that
geometric mean<=arithmetic mean
consider n+1 numbers where 1 is one of the numbers and the other n are 1+1/n
 
lurflurf said:
Use the fact that
geometric mean<=arithmetic mean
consider n+1 numbers where 1 is one of the numbers and the other n are 1+1/n

That only proves that it's monotonic.
 
Thanks for the neat proof.

one comment:

since we know that \sum 1/n! converges, we know e converges because
e = 1 + \sum 1/n! where n->infinity
 
Last edited:
jetplan said:
Thanks for the neat proof.

one comment:

since we know that \sum 1/n! converges, we know e converges because
e = 1 + \sum 1/n! where n->infinity

but first you'd have to prove that the original series converges and that its limit is equal to 1 + \sum_{n=1}^{\infty} 1/n! .
 
hamster143 said:
That only proves that it's monotonic.

2<(1+1/n)^n<(1+1/n)^(n+1)<4

By Monotone convergence theorem bounded monotonic series converge.
 
Last edited:
(1+1/n)^(n+1)<4

I don't see it.
 
^
(1+1/n)^(n+1) is decreasing thus
(1+1/n)^(n+1)<(1+1/1)^(1+1)=4
 
  • #10
You repeated use of "e is convergant" is confusing. Numbers do not converge. It makes sense to ask how to show that the limit \lim_{n\to\infty} (1+ 1/n)^n converges or how to show that \sum_{n=0}^\infty 1/n! converges.
 

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