# How to show that e converges ?

1. Aug 5, 2010

### jetplan

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

2. Aug 5, 2010

### hamster143

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.

3. Aug 5, 2010

### lurflurf

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

4. Aug 5, 2010

### hamster143

That only proves that it's monotonic.

5. Aug 5, 2010

### jetplan

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: Aug 5, 2010
6. Aug 5, 2010

### hamster143

but first you'd have to prove that the original series converges and that its limit is equal to 1 + $\sum_{n=1}^{\infty} 1/n!$ .

7. Aug 5, 2010

### lurflurf

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

By Monotone convergence theorem bounded monotonic series converge.

Last edited: Aug 5, 2010
8. Aug 5, 2010

### hamster143

I don't see it.

9. Aug 6, 2010

### lurflurf

^
(1+1/n)^(n+1) is decreasing thus
(1+1/n)^(n+1)<(1+1/1)^(1+1)=4

10. Aug 6, 2010

### HallsofIvy

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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