Rewriting an inequality

[armakrabbe]

I'm looking at chapter 1 problem 7 in the solution manual of Dan Stefanica's A Primer for the Mathematics of Financial Engineering. There's a step I don't quite understand, it's[math]x\ln(1+\frac{1}{x}) < 1 < (x + 1)\ln(1 + \frac{1}{x})[/math]which can be written as
[math]\frac{1}{x+1} < \ln(1 + \frac{1}{x}) < \frac{1}{x}[/math].
I don't understand what happened here and have been trying a bunch of simple operations to get from the first line to the second line. Any help on this would be much appreciated. Also, I was curious as to where I could ask for help for the book problems on this forum.

 

[Andy Nguyen]

Pinging the author @dstefan

 

[GWeitz]

Divide everything by ln(1+1/x) and invert everything

 

[Daniel Duffy]

An interestig follow-on is to prove the individual inequalities in step 1.

 

[Quasar Chunawala]

An interestig follow-on is to prove the individual inequalities in step 1.

I think that one part of the second inequality is to prove that :

[math]\left(1+\frac{1}{x}\right)^x < e[/math]
I can write the binomial expansion of the left-hand side, and of course the right hand side is [imath]\sum_{n=0}^{ \infty}\frac{1}{n!}[/imath]. I am not able to bound the left-hand side.

 

[Quasar Chunawala]

Pick an arbitrary but fixed [imath]x > 0[/imath]. We can bound the expression [imath]\left(1+\frac{1}{x}\right)^x[/imath] as:

[math]\left(1 + \frac{1}{x}\right)^x = 1 + x \cdot \frac{1}{x} + \frac{x(x-1)}{2!}\frac{1}{x^2}+\ldots+\frac{1}{x^x}[/math]
which implies that

[math]\left(1 + \frac{1}{x}\right)^x \leq 1 + |x| \cdot \frac{1}{x} + \frac{|x||x-1|}{2!}\frac{1}{x^2}+\ldots+\frac{1}{x^x}[/math]
Since [imath]|x - 1| \leq |x| + 1 < |x|[/imath] and in general [imath]|x -n| \leq |x| + n < |x|[/imath], it follows that:

[math]\left(1 + \frac{1}{x}\right)^x< 1 + \frac{|x|}{x}+\frac{1}{2!}\cdot\frac{|x|^2}{x^2}+\frac{1}{3!}\frac{|x|^3}{x^3}+\ldots+\frac{1}{x!}\cdot \frac{|x|^x}{x^x}[/math]
Since [imath]x > 0[/imath], [imath]|x|=x[/imath], so

[math]\left(1 + \frac{1}{x}\right)^x < \sum_{n=0}^x\frac{1}{n!} \leq \sum_{n=0}^{\infty}\frac{1}{n!} = e[/math]
for all [imath]x >0[/imath].

Of course, the domain of the definition is [imath](-\infty,-1)\cup(0,\infty)[/imath], so there has to be additional work.

 

[Daniel Duffy]

If x is real, then x! is probably wrong. But by trying you get more insights into related stuff.

Gamma function - Wikipedia

en.wikipedia.org

I think that there is an error in one of your inequalities.

what about asymptotic values of x? 0, -\infty, + \infty

Inequalities (Cambridge Mathematical Library) 2, Hardy, G. H., Littlewood, J. E., Pólya, G. - Amazon.com

Inequalities (Cambridge Mathematical Library) - Kindle edition by Hardy, G. H., Littlewood, J. E., Pólya, G.. Download it once and read it on your Kindle device, PC, phones or tablets. Use features like bookmarks, note taking and highlighting while reading Inequalities (Cambridge Mathematical...

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[Quasar Chunawala]

If x is real, then x! is probably wrong. But by trying you get more insights into related stuff.

Precisely.

Yeah, the proof attempt is pretty dodgy. Firstly, Newton's binomial expansion theorem is only applicable [imath]\forall n \in \mathbb{N}[/imath]. [imath](1+1/x)^{-n}[/imath] has a power-series representation.

I think the below would do the job for [imath]x > 1[/imath] (the interval of convergence of the infinite series [imath]\sum_{n=1}^{\infty}\frac{(-1)^n y^n}{n}[/imath] is [imath](-1,1][/imath]). Consider:

[math]\log\left(1+\frac{1}{x}\right) = \frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}-\frac{1}{4x^4}+\ldots[/math]
Rewriting as:

[math]\log\left(1+\frac{1}{x}\right) = \frac{1}{x}-\left[\left(\frac{1}{2x^2}-\frac{1}{3x^3}\right)+\left(\frac{1}{4x^4}-\frac{1}{5x^5}\right)+\ldots\right][/math]
Therefore,

[math]x\log\left(1+\frac{1}{x}\right) = 1-\left[\left(\frac{1}{2x}-\frac{1}{3x^2}\right)+\left(\frac{1}{4x^3}-\frac{1}{5x^4}\right)+\ldots\right][/math]
Now, each pair of terms in the round brackets are non-negative. Hence, [imath]x(1+1/x)^x[/imath] has upper bound [imath]1[/imath] (atleast for x > 1).

Also, [imath]\lim_{x \to \infty}\log(1+1/x)^x = 1[/imath].