Limits of Sequences .... Stoll Theorem 2.2.6

[Math Amateur]

In his book: Introduction to Real Analysis, Manfred Stoll does not prove parts (a) and (b) of Theorem 2.2.6 on the limits to certain special sequences ...

I am having trouble getting started on the proof of part (a) ... can someone please help me to make a meaningful start to the proof ... (despite the fact that Stoll informs us that the proof is straightforward ... :( ... )

Theorem 2.2.6 reads as follows:

View attachment 7194Peter

 

[Math Amateur]

In his book: Introduction to Real Analysis, Manfred Stoll does not prove parts (a) and (b) of Theorem 2.2.6 on the limits to certain special sequences ...

I am having trouble getting started on the proof of part (a) ... can someone please help me to make a meaningful start to the proof ... (despite the fact that Stoll informs us that the proof is straightforward ... :( ... )

Theorem 2.2.6 reads as follows:

Peter

After some reflection I have a proposed solution ...To show \(\displaystyle \text{lim}_{ n \rightarrow \infty } = 0\) Now ... for any \(\displaystyle \epsilon \gt 0\) the inequality \(\displaystyle \left\lvert\frac{1}{ n^p } - 0\right\rvert \lt \epsilon\)gives \(\displaystyle \frac{1}{ n^p } \lt \epsilon\) or \(\displaystyle n \gt \frac{1}{ \epsilon^{ \frac{1}{ p } }}\) ... ... Hence ... if we take \(\displaystyle N \gt \frac{1}{ \epsilon^{ \frac{1}{ p } } } \) ... then for \(\displaystyle n \gt N\) we have ...\(\displaystyle \frac{1}{ n^p } \lt \left(\frac{1}{ \frac{1}{ \epsilon^{ \frac{1}{p} } } }\right)^p = \epsilon \)Is that correct?

Peter

 

[Opalg]

After some reflection I have a proposed solution ...To show \(\displaystyle \text{lim}_{ n \rightarrow \infty } = 0\) Now ... for any \(\displaystyle \epsilon \gt 0\) the inequality \(\displaystyle \left\lvert\frac{1}{ n^p } - 0\right\rvert \lt \epsilon\)gives \(\displaystyle \frac{1}{ n^p } \lt \epsilon\) or \(\displaystyle n \gt \frac{1}{ \epsilon^{ \frac{1}{ p } }}\) ... ... Hence ... if we take \(\displaystyle N \gt \frac{1}{ \epsilon^{ \frac{1}{ p } } } \) ... then for \(\displaystyle n \gt N\) we have ...\(\displaystyle \frac{1}{ n^p } \lt \left(\frac{1}{ \frac{1}{ \epsilon^{ \frac{1}{p} } } }\right)^p = \epsilon \)Is that correct?

Peter

Yes! (Yes)

 

[Math Amateur]

Yes! (Yes)

Thanks Opalg ... it gives me great confidence when you or Evgeny confirm my effort ...

Thank you ...

Peter

 

[solakis1]

After some reflection I have a proposed solution ...To show \(\displaystyle \text{lim}_{ n \rightarrow \infty } = 0\) Now ... for any \(\displaystyle \epsilon \gt 0\) the inequality \(\displaystyle \left\lvert\frac{1}{ n^p } - 0\right\rvert \lt \epsilon\)gives \(\displaystyle \frac{1}{ n^p } \lt \epsilon\) or \(\displaystyle n \gt \frac{1}{ \epsilon^{ \frac{1}{ p } }}\) ... ... Hence ... if we take \(\displaystyle N \gt \frac{1}{ \epsilon^{ \frac{1}{ p } } } \) ... then for \(\displaystyle n \gt N\) we have ...\(\displaystyle \frac{1}{ n^p } \lt \left(\frac{1}{ \frac{1}{ \epsilon^{ \frac{1}{p} } } }\right)^p = \epsilon \)Is that correct?

Peter

In going from: \(\displaystyle \frac{1}{ n^p } \lt \epsilon\) to

\(\displaystyle n \gt \frac{1}{ \epsilon^{ \frac{1}{ p } }}\)

What are the axioms or theorems or definitions of the real Nos that you use??