Why does the sum S_n = 1 + 1/1! + 1/2! + ... + 1/n! satisfy S_n < 3?

[courtrigrad]

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

 

[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)

 

[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.

 

[ahrkron]

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".

 

[HallsofIvy]

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

 

[courtrigrad]

(3) is just a reference

 

[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.

 

[Muzza]

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

 

[dextercioby]

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

Daniel.

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

 

[Muzza]

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

Um, post #5?

 

[dextercioby]

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

Daniel.

 

[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) $$\lim_{x\rightarrow \infty} S_n = e$$.

 

[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.

 

[Muzza]

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

 

[cepheid]

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

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). ?

 

[courtrigrad]

sorry for any misunderstanding