# Number e

1. Jan 24, 2005

If $$S_n = 1 + \frac {1}{1!} + \frac {1}{2!} + ... + \frac {1}{n!}$$ [B}(1) [/B] and as n increases the sequence tends to a limit.

For all values of n $$S_n \leq 1 + 1 + \frac {1}{2} + \frac {1}{2^2} + ... + \frac {1}{2^{n-1}} = 1 + \frac {1 - \frac {1}{2^n}}{1 - \frac {1}{2}} < 3.$$ (2)

So $$\lim_{x\rightarrow \infty} S_n = e$$ (3)

What I do not understand is how they reached the conclusion in (2) ?

Any insight or ideas are appreciated

Thanks

Last edited: Jan 24, 2005
2. Jan 24, 2005

### fourier jr

the stuff with the 2^n s in the denominators is just a finite geometric series, and the last inequality on that line is just the formula for it. if you simplify that expression you get $$3-1/2^{n-1}$$ (that's what i got anyway)

3. Jan 24, 2005

### matt grime

The conclusion is a definition. They prove it (the sequence of partial sums) is an increasing sequence bounded above, hence converges to seomthing, which they are labelling e.

4. Jan 24, 2005

### ahrkron

Staff Emeritus
In step 2 they are:
1. Comparing two sums. The sum with factorials is, term by term, less or equal to the sum that has powers of 2 instead.
2. Obtaining an upper bound for the sum with powers of 2, by using the well known formula for a geometric sum.
3. Since (the sum with factorials) is less or equal than (the sum with powers of 2), which is less than 3, then the original sum is less than 3, hence it has an upper bound and a limit.

The third step in your post is basically saying "we now know the limit exists, let's call it e".

5. Jan 24, 2005

### HallsofIvy

And the last statement should be
$$\lim_{x\rightarrow \infty} S_n = e < 3$$

6. Jan 24, 2005

(3) is just a reference

7. Jan 24, 2005

### dextercioby

Nope,(3) should be a notation or a reversed definition.
$$\lim_{n\rightarrow +\infty} S_{n}\equiv^{notation} e$$

which means
$$e\equiv^{definition} \lim_{n\rightarrow +\infty}S_{n}$$

Daniel.

8. Jan 24, 2005

### Muzza

Surely courtigrad meant that the "(3)" itself was a reference.

9. Jan 24, 2005

### dextercioby

AAAAAAA...U mean number of an equation... :tongue2: That was obvious...Does he think we're blind or stupid??

Daniel.

P.S.Who did he reply to??? :tongue2: There's no context for his statement...

10. Jan 24, 2005

### Muzza

Um, post #5?

11. Jan 24, 2005

### dextercioby

Then he must be "ndfvjdfnvkjdfnvkjdfnvkfnv".Really.Do you see any connection between post #5 & his reply ??

Daniel.

12. Jan 24, 2005

### Muzza

Yes, I do. HallsofIvy thought that

$$\lim_{x\rightarrow \infty} S_n = e$$ (3)

meant

$$\lim_{x\rightarrow \infty} S_n = e < 3$$,

and then courtigrad pointed out that (3) was just meant as a reference to (the definition) $$\lim_{x\rightarrow \infty} S_n = e$$.

13. Jan 24, 2005

### dextercioby

I simply doubt HallsofIvy could have done something like that...It's not nice to speculate on what you think people think.In most of the cases it turns out you're wrong...

And HallsofIvy is HallsofIvy...

Daniel.

14. Jan 24, 2005

### Muzza

You know, sometimes I think you just enjoy being contrarian.

15. Jan 25, 2005

### cepheid

Staff Emeritus
Oh? I don't think Halls confused an equation label with an actual number in the statement. I thought he said there should be a < 3 in there because there was a < 3 in eqn (2). ???

16. Jan 25, 2005