MHB Apostal Chapter 4 - Cauchy Sequences - Example 1, Section 4.3, page 73

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I need some help in fully understanding Example 1, section 4.3 Cauchy Sequences, page 73 of Apostol, Mathematical Analysis.

Example 1, page 73 reads as follows:
https://www.physicsforums.com/attachments/3844
https://www.physicsforums.com/attachments/3845

In the above text, Apostol writes:

" ... ... If $$m \gt n \ge N$$ we find (by taking successive terms in pairs) that

$$| x_m - x_n |$$

$$= | \frac{1}{n+1} - \frac{1}{n+2} + \ ... \ \pm \frac{1}{m}| \lt \frac{1}{n} \le \frac{1}{N} $$

... ... "Can someone please explain to me exactly how we can show that

$$| x_m - x_n |$$

$$= | \frac{1}{n+1} - \frac{1}{n+2} + \ ... \ \pm \frac{1}{m}| \lt \frac{1}{n} \le \frac{1}{N} $$Help will be appreciated,

Peter
 
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Peter said:
I need some help in fully understanding Example 1, section 4.3 Cauchy Sequences, page 73 of Apostol, Mathematical Analysis.

In the above text, Apostol writes:

" ... ... If $$m \gt n \ge N$$ we find (by taking successive terms in pairs) that

$$| x_m - x_n |$$

$$= | \frac{1}{n+1} - \frac{1}{n+2} + \ ... \ \pm \frac{1}{m}| \lt \frac{1}{n} \le \frac{1}{N} $$

... ... "Can someone please explain to me exactly how we can show that

$$| x_m - x_n |$$

$$= | \frac{1}{n+1} - \frac{1}{n+2} + \ ... \ \pm \frac{1}{m}| \lt \frac{1}{n} \le \frac{1}{N} $$

For any n is...

$\displaystyle \frac {1}{n} > \frac{1}{n+1} > \frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3} > \frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3} - \frac{1}{n+4} + \frac{1}{n+5} > ...$

Kind regards

$\chi$ $\sigma$
 
chisigma said:
For any n is...

$\displaystyle \frac {1}{n} > \frac{1}{n+1} > \frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3} > \frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3} - \frac{1}{n+4} + \frac{1}{n+5} > ...$

Kind regards

$\chi$ $\sigma$
Thanks chisigma ... appreciate the help ...

BUT ... can you please help further ...

Why exactly is

$$\frac{1}{n+1} \gt \frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3}?$$

Can you help?

Petger
 
Peter said:
...Why exactly is

$$\frac{1}{n+1} \gt \frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3}?$$

Can you help?

Peter

Hello Peter! (Wave)

The above inequality is equivalent to (found by subtracting $$\frac{1}{n+1}$$ from both sides):

$$0\gt\frac{1}{n+3}-\frac{1}{n+2}$$

Or:

$$\frac{1}{n+2}\gt\frac{1}{n+3}$$
 
MarkFL said:
Hello Peter! (Wave)

The above inequality is equivalent to (found by subtracting $$\frac{1}{n+1}$$ from both sides):

$$0\gt\frac{1}{n+3}-\frac{1}{n+2}$$

Or:

$$\frac{1}{n+2}\gt\frac{1}{n+3}$$
Oh! Indeed!

Well, thanks Mark ...

Should have seen that ... :(

Thanks again,

Peter
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
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