Help Proving that the Sequence for the Natural Log Is Bounded

Date: 09/15/2009 at 09:17:24
From: Rahul
Subject: It is about the derivation of the number 'e'

Let a(sub n) be the sequence [1 + (1/n)]^n.

I understand that the sequence is monotonically increasing.  But in 
proving that it is also bounded, two of the steps remain unclear to 
me.

This step is OK for me:

    a(sub n) = 2 + (1/2!)[1 - (1/n)]
                 + (1/3!)[1 - (1/n)][1 - (2/n)]
                 + ... up to (n + 1) terms.

This step, however, I do not understand:

    a(sub n) < 2 + (1/2!) + (1/3!) + ...... + (1/n!)

The reasoning for this is given to be that

    1 - (1/n) < 1
    1 - (2/n) < 1

That is how those terms containing n in the brackets behave.  (The 
above inequalities also are understood because n being 1, 2, 3, ....
1/n  <= 1.)

Now, two times anything greater than 1 is greater than 2.  For example,
2*(1.5) = 3.

Two times anything less than 1 is less than 2.  For example,
2*(0.5) = 1; or 2*(0.9)= 1.8.

And 2*1 = 2.

So, the second step above is true for 2; but how do you generalize it?

Date: 09/16/2009 at 05:38:07
From: Doctor Ali
Subject: Re: It is about the derivation of the number 'e'

Hi Rahul!

Thanks for writing to Dr. Math.

Just note that when you accept that

     1 - (1/n) < 1
     1 - (2/n) < 1
     ...,

we can say that any product of the above expressions is still less
than one.

So, we can say that the original series are like

    1 + 1/2! + 1/3! + 1/4! + ...,

where every term is multiplied by a number less than one.  Therefore,
it is less than the above series.

Does that make sense?

Please write back if you still have any difficulties.

- Doctor Ali, The Math Forum
  http://mathforum.org/dr.math/