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

[Ed Quanta]

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?

 

[matt grime]

Why don't you work out the length of 1/n^x in the l_p norm.

 

[Ed Quanta]

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.

 

[matt grime]

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?

 

[Ed Quanta]

[ (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.

 

[matt grime]

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.

 

[Ed Quanta]

It isn't clear to me where p = infinity

 

[matt grime]

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.