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just.so
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[SOLVED] Sequences in lp spaces... (Functional Analysis)
Find a sequence which converges to zero but is not in any lp space where 1<=p<infinity.
N/A
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!
Justin
Homework Statement
Find a sequence which converges to zero but is not in any lp space where 1<=p<infinity.
Homework Equations
N/A
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!
Justin
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