Can the Mean Value Theorem be Used to Prove Inequalities for Natural Numbers?

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Homework Help Overview

The discussion revolves around proving an inequality involving the natural logarithm for all natural numbers, specifically the inequality \( \frac{1}{x+1} < \ln(x+1) - \ln(x) < \frac{1}{x} \). The context is related to the Mean Value Theorem and its application in this scenario.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the use of the Mean Value Theorem and consider rewriting the logarithmic expression. There is a suggestion to use proof by induction instead of the Mean Value Theorem. Questions arise regarding the applicability of the theorem to natural numbers as a subset of real numbers.

Discussion Status

Participants are actively discussing various approaches, including the potential use of induction and the Mean Value Theorem. Some guidance has been offered regarding rewriting the logarithmic expression, but there is no explicit consensus on the best method to proceed.

Contextual Notes

There is a question about whether the Mean Value Theorem can be applied to natural numbers, as well as considerations about the assumptions regarding the nature of \( x \) being a positive integer.

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Homework Statement


I need to prove the following for all natural numbers:

1/(x+1) < ln(x+1) - ln(x) < 1/x


Homework Equations





The Attempt at a Solution


It's in the part of the mean value theorem problems, I try using it didn't go any where. I tried thinking of other ways, but nothing seems to work.
 
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The first thing I would do is write ln(x+1)- ln(x) as ln((x+1)/x). Are you saying that x must be a positive integer? Then I would assume that you should use proof by induction, not the mean value theorem!
 
Induction would work nicely, but taking advantage of your section hint try
log(x+1)-log(x)=[log(x+1)-log(x)]/[(x+1)-x]=log'(x+t)=1/(x+t)
where t is some number such that 0<t<1
so the problem is reduced to showing that
when ever 0<t<1
1/(x+1)<1/(x+t)<1/(x+0)
 
I got it with the mean value theorem, but because that's for all real numbers, can I say it's true for natural numbers too because it is a subset of the real numbers? Thanks for the help.
 

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