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

In summary, the conversation discusses an inequality in Example 1, section 4.3 of Apostol's Mathematical Analysis regarding Cauchy sequences. The inequality is shown to be true using a series of fractions, and the conversation continues with further explanations and clarifications.
  • #1
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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 \(\displaystyle m \gt n \ge N\) we find (by taking successive terms in pairs) that

\(\displaystyle | x_m - x_n |\)

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

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

\(\displaystyle | x_m - x_n |\)

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

Peter
 
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  • #2
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 \(\displaystyle m \gt n \ge N\) we find (by taking successive terms in pairs) that

\(\displaystyle | x_m - x_n |\)

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

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

\(\displaystyle | x_m - x_n |\)

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

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

$\chi$ $\sigma$
 
  • #3
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

\(\displaystyle \frac{1}{n+1} \gt \frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3}?\)

Can you help?

Petger
 
  • #4
Peter said:
...Why exactly is

\(\displaystyle \frac{1}{n+1} \gt \frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3}?\)

Can you help?

Peter

Hello Peter! (Wave)

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

\(\displaystyle 0\gt\frac{1}{n+3}-\frac{1}{n+2}\)

Or:

\(\displaystyle \frac{1}{n+2}\gt\frac{1}{n+3}\)
 
  • #5
MarkFL said:
Hello Peter! (Wave)

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

\(\displaystyle 0\gt\frac{1}{n+3}-\frac{1}{n+2}\)

Or:

\(\displaystyle \frac{1}{n+2}\gt\frac{1}{n+3}\)
Oh! Indeed!

Well, thanks Mark ...

Should have seen that ... :(

Thanks again,

Peter
 

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

What is a Cauchy sequence?

A Cauchy sequence is a sequence of real numbers in which the terms become arbitrarily close to each other as the sequence progresses. In other words, for any given positive real number, there exists a point in the sequence after which all subsequent terms are within that distance from each other.

How is a Cauchy sequence different from a convergent sequence?

A Cauchy sequence is different from a convergent sequence in that a Cauchy sequence only requires that the terms become arbitrarily close to each other, while a convergent sequence requires that the terms approach a specific limit as the sequence progresses.

What is the significance of Example 1 in Section 4.3 of Apostal Chapter 4?

Example 1 serves as an illustration of how to determine if a given sequence is a Cauchy sequence. It also highlights the importance of the Cauchy criterion, which states that a sequence is Cauchy if and only if for any given positive real number, there exists a point in the sequence after which all subsequent terms are within that distance from each other.

How can we prove that a sequence is Cauchy?

To prove that a sequence is Cauchy, we can use the Cauchy criterion mentioned above. This involves showing that for any given positive real number, there exists a point in the sequence after which all subsequent terms are within that distance from each other. We can also use other techniques such as the squeeze theorem or the definition of convergence to prove that a sequence is Cauchy.

Can a sequence be both Cauchy and divergent?

No, a sequence cannot be both Cauchy and divergent. This is because a Cauchy sequence requires that the terms become arbitrarily close to each other, while a divergent sequence does not have a limit and the terms do not approach a specific value. Therefore, a sequence cannot satisfy both criteria at the same time.

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