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Use the defintion of a logarithmic function to prove $$ \frac{1}{x+1} \le \ln(1+\frac{1}{x}) \le \frac{1}{x}$$ for any positive integer x. ( Using defintion : $e^y=\ln^{-1}y$ )

Attempt: Using the definition, $e^y=\ln^{-1}y$ , I tried to prove the right hand side of the equality first by $ 1 +\frac{1}{x} \gt \frac{1}{x}$,
$\ln(1 +\frac{1}{x})\lt \frac{1}{x} \Rightarrow e^{\ln(\frac{x+1}{x})} \le e^{\frac{1}{x}}$ but I am really stuck and lost on how I should do it so that one of the sides there is ln and the other side is the value and also for the equality to be $\le$ as $ 1 +\frac{1}{x} \gt \frac{1}{x}$ is strictly less than and is not equal.

dormacus
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  • @user236182 , that question uses MVT – dormacus Nov 04 '17 at 14:47
  • @mechanodroid uses integration as a definition . – dormacus Nov 04 '17 at 14:47
  • There are a lot of linked duplicate questions in the link I gave. – user236182 Nov 04 '17 at 14:50
  • Now How do you define $e^x$. or $\ln x?$ – Guy Fsone Nov 04 '17 at 14:51
  • @dormacus I posted a new answer there. – user236182 Nov 04 '17 at 14:56
  • Let me be clear between $e^x $ and $\ln x$ has to be define first without mentioning the other one. Then WHAT IS YOUR DEFINITION OF $\ln x$ without using $e^x$. Since you use $\ln x$ to define $e^x$. you must give us the definition of $\ln x.$ – Guy Fsone Nov 04 '17 at 15:36
  • For any $x>0$ We have $$\frac{x}{x+1} =\int_{0}^x\frac{dt}{x+1} \le \int_{0}^x\frac{dt}{t+1} =\color{red}{\ln(x+1 )}=\int_{0}^x\frac{dt}{t+1} \le \int_{0}^x\frac{dt}{1} = x $$

    that is

    $$\frac{x}{x+1} \le \ln(x+1 ) \le x $$

    taking $X=\frac1x$ you get

    $$\frac{1}{X+1} \le \ln(\frac1X+1 ) \le \frac1X $$

     – Guy Fsone Nov 27 '17 at 15:53

1 Answers1

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Partial answer using log defined only as the inverse of the exponential : For the second inequality, use the property of the exponential function that for all real y, $$e^y \geq 1+y$$ Take log on both sides (here is where the log as inverse of the exponential comes in) and use $y=1/x$ to get the result.

Mathemagical
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  • This answer is more in line with the OP’s question. How is the inequality on exponential proven? – Zach Boyd Nov 04 '17 at 15:08
  • Two ways for positive y: 1) consider the series expansion of the exponential. All the terms after 1+ y are positive. That gives the inequality. The other way is to note that the exponential as well as 1+y have the same value and the same slope at y=0. However, the slope of 1+y is always 1 while the slope exponential is greater than 1 for any positive y (and keeps increasing). – Mathemagical Nov 04 '17 at 15:46
  • Do you know a way without calculus? It seems like an elementary proof should be possible – Zach Boyd Nov 04 '17 at 15:48
  • One can use the definition of the e as the limit as $y \longrightarrow 0$ of $(1+y)^{\frac{1}{y}}$. e is the supremum over all positive y values of this function (increasing as y approaches 0 from the right). Since $$(1+y)^{\frac{1}{y}} \leq e$$, raising both sides to the power of y gives the inequality. – Mathemagical Nov 04 '17 at 16:04
  • @ZachBoyd See https://math.stackexchange.com/questions/504663/simplest-or-nicest-proof-that-1x-le-ex and linked questions there on the right. – user236182 Nov 04 '17 at 19:58