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

[Math Amateur]

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

 

[chisigma]

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

$\chi$ $\sigma$

 

[Math Amateur]

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

 

[MarkFL]

...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:

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

 

[Math Amateur]

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