Is {1/n^x} an Element of l^p for Various Values of x and p?

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Discussion Overview

The discussion centers on the conditions under which the sequence {1/n^x} belongs to the space l^p for various values of x and p. Participants explore convergence criteria and relationships between different l^p spaces, including implications for specific values of x and p.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that for x=1, {1/n^x} is in l^p for p>=2, based on the convergence of the series.
  • Another participant questions the convergence of the series 1/n^x for x>1 and proposes that if it converges, then {1/n^x} would be in l^p for p>x.
  • A different participant states that the sum of 1/n^r converges if and only if r>1, implying a relationship with the l^p norm.
  • One participant presents the l^p norm expression for the series and asserts that convergence occurs only where p>x.
  • There is a query about the subset relationship between l^p and l^q for p
  • Another participant clarifies that a sequence is in l^p if the sum of its terms raised to the power p converges, leading to the condition px>1.
  • Concerns are raised about the case when p=∞, with a participant seeking clarification on the conditions for membership in l^∞.
  • One participant explains that a sequence is in l^∞ if its terms are bounded, relating this to the earlier discussions on convergence.

Areas of Agreement / Disagreement

Participants express differing views on the convergence of the series {1/n^x} for various x and p, with no consensus reached on the implications of these conditions. Some participants agree on certain convergence criteria, while others challenge or seek clarification on specific cases.

Contextual Notes

The discussion includes assumptions about convergence and the definitions of l^p spaces that may not be universally agreed upon. The implications of the relationships between different values of p and x remain unresolved.

Ed Quanta
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So if we let x>0, For which 0<p<=infinity is {1/n^x} an element of l^p?

If x=1, then 1/n^x is clearly an element of l^p for p>=2, since for all these vector spaces, the series of 1/n will converge?

But if x<1, then in it seems that only for p=infinity, will {1/n^x} be an element of l^p. Is this correct?
 
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Why don't you work out the length of 1/n^x in the l_p norm.
 
The series 1 +1/2+1/3+...+1/n doesn't converge. Now if we have the summation of 1/n^x, and x>1, will 1 +1/2^x + 1/3^x + ... + 1/n^x converge? If this does always converge, then as long as p>x the series of {1/n^x} will be an element of l^p,right? That is what I meant to ask before.
 
the sum of 1/n^r converges if and only if r>1. HOpefully you can use that with the l^p norm: what is the l^p norm of the series 1/n^x?
 
[ (1/1)^p + (1/2^x)^p +...+ (1/n^x)^p]^(1/p)

And as I stated in my previous post, this should only converge where p>x unless I am missing something.


One more question. If p<q, then why must l^p be a subset of l^q? I know that there are two cases to prove here. One in which q is finite, and one in which q is infinity. But I am not sure how I can show that this is true.
 
If you think all the series are finite, then yes, you are missing something.

I think you missing something else too.

(a_1,a_2,...) is in l^p iff the sum of (a_i)^p converges, right?

so a_r = 1/r^x is in l^p iff the sum of (1/r^x)^p converges, which is iff px>1.

Right? Or are we talking at cross purposes? It should be clear now why the containment you mention is true.
 
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It isn't clear to me where p = infinity
 
What isn't clear to you? (a_1,a_2,...) is in L^{infinity} if and only if its terms are bounded, so 1/n^x is in it iff x>0 (which agrees with the notion of px>1 as it happens, as p tends to infinity.

Or do you mean the containment?

If a sum converges its terms tend to zero, and in particular are bounded so its l infinity.
 

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