Show that this is less than that

[ArcanaNoir]

Homework Statement

I would like to prove that when x is between 1 and 2 $$\frac{k^{0.5}x^x}{1+x^{2k}} < \frac{1}{\sqrt{x-1}}$$ for all $k\in \mathbb{N}$

Homework Equations

The Attempt at a Solution

This is for my analysis class. I am algebraist, so please take pity :)
I have learned that if I were to show that the numerator of the first fraction is less than the numerator of the second and the denominator of the first fraction is greater than the denominator of the second then the first fraction would be smaller than the second. That technique doesn't seem to be working though. If the second fraction needs tweaking to make this work then that is fine. I'm just trying to bound the first fraction with something integrable. I think my bound works based on Maple graphs.

 

[LCKurtz]

Does the bound have to be independent of $k$?

 

[ArcanaNoir]

Does the bound have to be independent of $k$?

Yes it does.

 

[dirk_mec1]

Can't be true, if I let k-> inf surely it will not be valid.

 

[ArcanaNoir]

Can't be true, if I let k-> inf surely it will not be valid.

alas! you may be right. I have just received word that a classmate may have a proof that the sequence cannot be bounded by an integrable function.

 

[D H]

Can't be true, if I let k-> inf surely it will not be valid.

Sure it is. The left hand side goes to zero as k→∞ for all x in (1,∞), and 0<1/sqrt(x-1) for all x in (1,2).

 

[dirk_mec1]

Yes you're right. Then how will the proof read?

 

[I like Serena]

Homework Statement

I would like to prove that when x is between 1 and 2 $$\frac{k^{0.5}x^x}{1+x^{2k}} < \frac{1}{\sqrt{x-1}}$$ for all $k\in \mathbb{N}$

Homework Equations

The Attempt at a Solution

This is for my analysis class. I am algebraist, so please take pity :)
I have learned that if I were to show that the numerator of the first fraction is less than the numerator of the second and the denominator of the first fraction is greater than the denominator of the second then the first fraction would be smaller than the second. That technique doesn't seem to be working though. If the second fraction needs tweaking to make this work then that is fine. I'm just trying to bound the first fraction with something integrable. I think my bound works based on Maple graphs.

Here's a possible outline of a proof.

Take the natural logarithm of the left hand side.
\begin{aligned}\ln \frac{k^{0.5}x^x}{1+x^{2k}}
&= 0.5\ln k + x\ln x - \ln(1+x^{2k}) \\
&< 0.5\ln k + x\ln x - \ln(x^{2k}) \\
&= 0.5\ln k + x\ln x - 2k \ln x
\end{aligned}
Find the maximum value of this last expression with respect to $k$, treating it as a real number.
The result is $k=1/\ln x$ with the maximum $\frac 1 2 \ln(1/\ln x) - 1$.

Therefore an upper bound of the original left hand side is:
$$\frac 1 {e\sqrt{\ln x}}$$

What is left is the proof that this is less than the right hand side.
Substitute $x=u+1$ and make a first order Taylor expansion of $\ln$.

And... we're done.

 

[ArcanaNoir]

Thank you so much ILS. This analysis makes me want to cry!

 

[I like Serena]

Thank you so much ILS. This analysis makes me want to cry!

Cry? Why?