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/
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