MHB Limits of Sequences .... Stoll Theorem 2.2.6

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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
 
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Peter said:
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 $$\text{lim}_{ n \rightarrow \infty } = 0$$ Now ... for any $$\epsilon \gt 0$$ the inequality $$\left\lvert\frac{1}{ n^p } - 0\right\rvert \lt \epsilon$$gives $$\frac{1}{ n^p } \lt \epsilon$$ or $$n \gt \frac{1}{ \epsilon^{ \frac{1}{ p } }}$$ ... ... Hence ... if we take $$N \gt \frac{1}{ \epsilon^{ \frac{1}{ p } } } $$ ... then for $$n \gt N$$ we have ...$$\frac{1}{ n^p } \lt \left(\frac{1}{ \frac{1}{ \epsilon^{ \frac{1}{p} } } }\right)^p = \epsilon $$Is that correct?

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

Peter
Yes! (Yes)
 
Opalg said:
Yes! (Yes)
Thanks Opalg ... it gives me great confidence when you or Evgeny confirm my effort ...

Thank you ...

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

Peter

In going from: $$\frac{1}{ n^p } \lt \epsilon$$ to

$$n \gt \frac{1}{ \epsilon^{ \frac{1}{ p } }}$$

What are the axioms or theorems or definitions of the real Nos that you use??
 
I posted this question on math-stackexchange but apparently I asked something stupid and I was downvoted. I still don't have an answer to my question so I hope someone in here can help me or at least explain me why I am asking something stupid. I started studying Complex Analysis and came upon the following theorem which is a direct consequence of the Cauchy-Goursat theorem: Let ##f:D\to\mathbb{C}## be an anlytic function over a simply connected region ##D##. If ##a## and ##z## are part of...
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