MHB Limits of Sequences .... Bartle and Shebert, Example 3.4.3 (b) ....

[Math Amateur]

I am reading "Introduction to Real Analysis" (Fourth Edition) b Robert G Bartle and Donald R Sherbert ...

I am focused on Chapter 2: Sequences and Series ...

I need help in fully understanding Example 3.4.3 (b) ...Example 3.4.3 (b) ... reads as follows:https://www.physicsforums.com/attachments/7230
In the above text from Bartle and Sherbert we read the following:

" ... ... Note that if $$z_n := c^{ \frac{1}{n} }$$ then $$z_n \gt 1$$ and $$z_{ n+1 } \lt z_n$$ for all $$n \in \mathbb{N}$$. (Why?) ... "Can someone help me to show rigorously that $$z_n \gt 1$$ and $$z_{ n+1 } \lt z_n$$ for all $$n \in \mathbb{N}$$ ... ... ?Hope that someone can help ...

Peter

 

[Opalg]

In the above text from Bartle and Sherbert we read the following:

" ... ... Note that if $$z_n := c^{ \frac{1}{n} }$$ then $$z_n \gt 1$$ and $$z_{ n+1 } \lt z_n$$ for all $$n \in \mathbb{N}$$. (Why?) ... "Can someone help me to show rigorously that $$z_n \gt 1$$ and $$z_{ n+1 } \lt z_n$$ for all $$n \in \mathbb{N}$$ ... ... ?

If $c>1$ then $c^n < c^{n+1}$. Raise both sides to the power $\frac1{n(n+1)}$ to get $c^{1/(n+1)} < c^{1/n}$. In other words, $z_{n+1} < z_n.$

 

[Math Amateur]

If $c>1$ then $c^n < c^{n+1}$. Raise both sides to the power $\frac1{n(n+1)}$ to get $c^{1/(n+1)} < c^{1/n}$. In other words, $z_{n+1} < z_n.$

Thanks Opalg ... but ... can you indicate how we prove that $$z_n \gt 1$$ ... ?

Peter

 

[Opalg]

Thanks Opalg ... but ... can you indicate how we prove that $$z_n \gt 1$$ ... ?

Peter

Suppose that $z_n\leqslant1$. Then $z_n^2 \leqslant z_n \leqslant 1$, $z_n^3 \leqslant z_n^2 \leqslant 1$, ... , and by induction $z_n^k\leqslant 1$ for all $k\in\Bbb{N}.$ In particular, $z_n^n = c \leqslant1$. But that contradicts the fact that $c>1$. So the initial supposition was wrong, and therefore $z_n>1.$

 

[solakis1]

Thanks Opalg ... but ... can you indicate how we prove that $$z_n \gt 1$$ ... ?

Peter

$$c>1\Longrightarrow lnc>ln1\Longrightarrow lnc>0\Longrightarrow\frac{1}{n}lnc>0\Longrightarrow lnc^{\frac{1}{n}}>ln1\Longrightarrow c^{\frac{1}{n}}>1 $$

 

[Opalg]

$$c>1\Longrightarrow lnc>ln1\Longrightarrow lnc>0\Longrightarrow\frac{1}{n}lnc>0\Longrightarrow lnc^{\frac{1}{n}}>ln1\Longrightarrow c^{\frac{1}{n}}>1 $$

This problem comes from the book by Bartle and Sherbert, which aims to develop real analysis rigorously from first principles. At this stage of the book the logarithm function has not yet been introduced, so it is not available for use.

 

[solakis1]

This problem comes from the book by Bartle and Sherbert, which aims to develop real analysis rigorously from first principles. At this stage of the book the logarithm function has not yet been introduced, so it is not available for use.

In another thread the book asks us in exercise 2.1.12(1) to prove :

$$a>0\Longrightarrow 0<\frac{a}{2}<a$$

To prove that we need the definition 1+1=2

Does the book mention that definition anywhere ?

If the answer is yes then i stop the discussion here

But if the answer is no, then one may wonder how rigorously the book aims to develop real analysis.

However one may say ,but 1+1=2 is so well known that the book may omit such obvious definition.

In the same way lun function is so well known even in high school that we may use it,
although the book examines it in more details in later sections