Prove that e-(1+1/1!+1/2!+1/3!+.......+1/n!)<1/(n!n) for all natural number n.

Hence prove that e is an irrational number !

Anyone could help ?
Thaks

>Prove that e-(1+1/1!+1/2!+1/3!+.......+1/n!)<1/(n!n) for all
>natural number n.
>
>Hence prove that e is an irrational number !
>

Assume e is rational M/N in lower terms. Then

M/N - (1+1/1!+1/2!+1/3!+.......+1/N!)<1/(N!N), or

M·N!/N - (N!+N!/1!+N!/2!+N!/3!+.......+N!/N!)<1/N.

M·N!/N is a whole number, as is the the number in parentheses. Since their difference is an integer, the inequality implies that the two are equal:

M·N!/N = N!+N!/1!+N!/2!+N!/3!+.......+N!/N!, or

e = 1+1/1!+1/2!+1/3!+.......+1/N!,

which contradicts the fact that

e = 1+1/1!+1/2!+1/3!+.......+1/N! + a little something.

Therefore, our assumption that e = M/N, leads to a contradiction.

I had some trouble with the first part:

Prove that e - (1+1/1!+1/2!+...+1/n!) < 1/(n!n) for all n in {1,2,...}

It was pretty easy to prove that the residual is less than (e-1)/(n+1)!, but that's not good enough. After a little trial and error, I found the following manipulations. (Of course I discovered the transformation starting at the bottom and working my way to the top.)

{n/(2n+6)}{1 + 1/(n+4) + (n+3)!/(n+5)! + (n+3)!/(n+6)! + ...} < (1/2)(e-1) < 1

The first inequality follows because

n/(2n+6) < 1/2

and

1 + 1/(n+4) + (n+3)!/(n+5)! + (n+3)!/(n+6)! + ... < 1 + 1/2 + 1/3! + 1/4! + ... = e-1

So, if we are prepared to accept the fact that (e-1)/2 < 1, then we can proceed...

{1/(n+3)}{1 + 1/(n+4) + (n+3)!/(n+4)! + ...} < 2/n

1/(n+3) + (n+2)!/(n+3)! + (n+2)!/(n+4)! + ... < 2/n

1 + 1/(n+3) + (n+2)!/(n+3)! + (n+2)!/(n+4)! + ... < 1 + 2/n = (n+2)/n

1/(n+2) + (n+1)!/(n+3)! + (n+1)!/(n+4)! + ... < 1/n

1 + 1/(n+2) + (n+1)!/(n+3)! + (n+1)!/(n+4)! + ... < (n+1)/n

{n/(n+1)}{1 + 1/(n+2) + (n+1)!/(n+3)! + ...} < 1

Therefore,

e - (1 + 1/1! + 1/2! + 1/3! + ... + 1/n!)
= (1/(n+1)! + 1/(n+2)! + ...)
= (1/n!n){n/(n+1)}{1 + 1/(n+2) + (n+1)!/(n+3)! + ...}
< (1/n!n)

Q.E.D.

Does it matter if e is irrational in order to solve to problem?

Heres another way to prove...

e - (1+1/1!+1/2!+...+1/n!) < 1/(n!n)

Well, e=1+1/1!+1/2!+..., so
e - (1+1/1!+1/2!+...+1/n!)=1/(n+1)!+1/(n+2)!+1/(n+3)!...
1/(n+1)!+1/(n+2)!+1/(n+3)!... is also equal to...
1/n!*1/(n+1)+1/n!*1/((n+1)(n+2))
in other words, if n=6 then we have...
1/6!*1/7+1/6!*1/(7*8)+1/6!*1/(7*8*9)....
And we need to prove this is less than
1/(n!n)
Which if n=6, then thats like 1/6!*1/6
A 1/6! can be factored out of both sides of the equation so, so far that we have...
1/7+1/(7*8)+1/(7*8*9)....<1/6
Now take a geometric series where the change is 1/7, it would look like..
1/7+1/(7*7)+1/(7*7*7)... we kind find the sum of all these with the common equation...
starting # / (1 - change)
(1/7)/(1-1/7)=1/6. Infact, for all sequences like this, they add up to 1/(n-1)
So, if the sequecnce with the 7's add up to 1/6, then the sequence 1/(7)+1/(7*8)+1/(7*8*9)... must be smaller than 1/6 because the numbers on the bottom get bigger than 7

e - (1+1/1!+1/2!+...+1/n!) < 1/(n!(n+1))
If this were the equation (notice the difference at the end), it would always be false because...
1/(n+1) will always be less than 1/(n+1)+some other stuff...

If it looked like this....
e - (1+1/1!+1/2!+...+1/n!) < 1/(n!(n+.5))
I think it would start to depend upon what n was.

Your proof relies on the Calculus of Taylor Polynomials. While this is perfectly valid, I'm not sure this method is allowed in the class that posed the problem. If you were to use calculus, you'd have to:
1) Find the T.P. of e^x (which is 1/0! + 1/1! + 1/2! ... + 1/n!)
2) Use the error formula and its approximations, and knowledge that e < 3 (if you can find that out) to find a reasonable way to squeeze e between two inequalities.
3) You'd pick an n to which you'd take the T.P. and multiply the inequalities through by n!

Then it will follow that n!(e - T.P.(e)) is an integer, but it is greater than 0 and less than 3/4, which is impossible. Thus e cannot be rational.

I think some of the earlier replies are actually good, if carefully checked.