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In "Mathematical Methods for Physicists: A Comprehensive Guide, " Seventh edition, by G. B. Arfken, H. J. Weber and F. E. Harris question 1.2.13 asks the reader to show, with $n>1\,,$ that $$ \mbox{(a)}\,\,\,\, \frac{1}{n} \,\,\,\,-\,\,\,\, \ln\left(\frac{n}{n-1}\right)\,\,\,\, <\,\,\,\, 0\,,\quad\,\,\quad \mbox{(b)}\,\,\,\,\, \frac{1}{n} \,\,\,\,-\,\,\,\, \ln\left(\frac{n+1}{n}\right) \,\,\,\,\,>\,\,\,\,0\,. $$

I have attempted to construct some proofs of the given inequalities, which I shall now proceed to give. Any feedback on the validity of their construction would be welcome. (I ask because I am not a mathematician.)

(a) We know that $$\exp(x) > 1 + x\,, \forall\,\, x\in \mathbb{R}_{\ne 0}\,.$$ Restricting the values of $x$ to satisfy $x < 0$, and letting $x= \ln(n-1) - \ln(n)$ and requiring $n>1$ we have $$ \exp\left[\ln\left(\frac{n-1}{n}\right)\right]\,\,\,\, > \,\,\,\, 1 \,\,\,\, + \,\,\,\,\ln\left(\frac{n-1}{n}\right)\,. $$ Thus $$ \frac{n-1}{n}\,\,\, > \,\,\,\,\,1 \,\,\,+ \,\,\,\ln\left(\frac{n-1}{n}\right)\,.$$ Therefore $$ 0\,\,\, >\,\,\,\, \frac{1}{n} - \,\,\ln\left(\frac{n}{n-1}\right)\,.$$

(b) Again we use the fact that $$\exp(x) > 1 + x\,, \forall\,\, x\in \mathbb{R}_{\ne 0}\,.$$ This time the values of $x$ are restricted to satisfy $x>0\,.$ Letting $x=\dfrac{1}{n}$ and requiring $n>1$ we have $$ \exp\left(\frac{1}{n}\right)\,\,\,\,\,>\,\,\,\,\,1\,\,\,\,\,+\,\,\,\,\,\frac{1}{n}\,. $$ Taking the natural logarithm of both sides of this inequality it is found that $$ \ln\left[\exp\left(\frac{1}{n}\right)\right]\,\,\,\,\,\,>\,\,\,\,\,\,\,\ln\left(1\,\,+\,\,\frac{1}{n}\right)\,. $$ Thus $$ \frac{1}{n}\,\,\,\,>\,\,\,\,\ln\left(1+\frac{1}{n}\right)\,. $$ Therefore $$ \frac{1}{n}\,\,\,\,-\,\,\,\,\ln\left(\frac{n+1}{n}\right) > 0\,. $$

Edit: My main concern about the validity of the above proofs is that I am worried that may be $x$ should have the same dependency on $n$ in parts (a) and (b). Should I be concerned or are the proofs okay? If $x$ were to have the same dependency on $n$ in each case wouldn't a different inequality have to be used to begin with in each part?

Physics_Et_Al
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1 Answers1

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Your proofs are fine. In a sense, both inequalities you proved are 'special cases' of the general inequality you started with. Since they are different special cases, it makes sense that you need to plug in different values for $x$ to derive them.

In other words, from the general inequality you started with, both $e^{0.1} > 1.1$ and $e^{-0.1} > 0.9$ follow. It doesn't make those statements any less true that you needed to plug in different values for $x$ to obtain them.

Ingix
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