# Homework Help: Sequences in lp spaces (Functional Analysis)

1. Mar 2, 2008

### just.so

[SOLVED] Sequences in lp spaces... (Functional Analysis)

1. The problem statement, all variables and given/known data
Find a sequence which converges to zero but is not in any lp space where 1<=p<infinity.

2. Relevant equations
N/A

3. The attempt at a solution
I strongly suspect 1/ln(n+1) is a solution.

Since ln(n+1) -> infinity as n -> infinity, we have 1/ln(n+1) -> 0 as n-> infinity.

I attempted using the integration test to show that partial sums for the sequence (1/ln(n+1))^p do not converge, but integrating 1/ln(n) became a bit problematic!

Attempt two was to show that each successive term in the sequence {(1/ln(n+1))^p} is larger than each successive term in some tail of the harmonic sequence.

i.e. There exists m such that
(1/ln(2))^p >= 1/(m+0) or m >= (ln(2))^p
(1/ln(3))^p >= 1/(m+1) or m >= (ln(3))^p - 1
(1/ln(4))^p >= 1/(m+2) or m >= (ln(4))^p - 2
.
.
.
In general for such an m to exist, m >= (ln(x+2))^p - x

And therefore I need to show that this right hand side is bounded above. I suspect this is true, and with a lot of hand waving can convince myself that the "ln" part of the RHS is "stronger" than the "^p" part and thus (ln(x+2))^p will eventually "slow down enough" so as to be less than x for large enough values of x.

But my log work leaves a little to be desired and I can't even prove there is a solution to the equation (ln(x+2))^p = x.

Any help much appreciated!

Best regards,
Justin

Last edited: Mar 2, 2008
2. Mar 2, 2008

### quasar987

Never mind. :x

3. Mar 2, 2008

### quasar987

There is a sweet test when it comes to series with logarithm in the argument. It says if a sequence a_n is decreasing and non-negative, then the series of a_n converges if and only if the series of 2^n*a_{2^n} does.

So here,

$$a_n=\frac{1}{\ln(n)^p}$$

and so

$$2^na_{2^n} = \frac{2^n}{n^p\ln(2)^p}$$

It's called the Cauchy condensation test: http://en.wikipedia.org/wiki/Cauchy_condensation_test

Last edited: Mar 2, 2008
4. Mar 2, 2008

### just.so

That IS sweet!

A gazillion thanks!

J