I Behavior of a function for large x?

[pellman]

TL;DR Summary

Need to learn how to analyze the asymptotic behavior of functions

I have a problem asking to show that a certain function approaches a quadratic for large values of the variable. And I realize now that this is a skill with which I am totally unfamiliar. Can't use a Taylor series in y= 1/x because the value at y=0 is infinite. Would appreciate a recommended resource that covers this topic.

I would like to post the specific problem but I don't know how to post equations here. Is there a how-to page?

 

[fresh_42]

A bit more details would be fine. Maybe you could expand it into a Taylor series at another value.

I would like to post the specific problem but I don't know how to post equations here. Is there a how-to page?

https://www.physicsforums.com/help/latexhelp/

 

[pellman]

The specific problem is to show that

$x \sum_{p=1} ^\infty \ln \left( \frac 1 { 1 - e^{-p/x}} \right)$

approaches

$\frac { \pi^2} {6} x^2$

for large x. So we could drop the leading x and just show that the sum is linear for large x. I know that the pi^2 / 6 comes from $\sum_{p=1}^\infty \frac 1 {p^2}$

 

[fresh_42]

So we have to show
\begin{align*}
\zeta(2)x&\sim\sum_{p=1}^\infty \ln\left(\dfrac{1}{1-e^{-p/x}}\right) \\
&= -\ln\left( \prod_{p=1}^\infty (1-e^{-p/x})\right)\\
&\Longleftrightarrow\\
e^{\zeta(2)x}&\sim\prod_{p=1}^\infty \dfrac{1}{1-e^{-p/x}}
\end{align*}
and $\zeta(2)=\displaystyle{\dfrac{\pi^2}{6}=\prod_{p\text{ prime }}\dfrac{1}{1-p^{-2}}=\sum_{p=1}^\infty \dfrac{1}{p^2}}$

I would start with the Taylor series for $e^{-p/x}$ and try to get an asymptotic behavior that goes with $1/p^2$ in the summation.

If this shouldn't work, I would look for formulas and integration tricks in the realm of the zeta function. There are so many formulas around, that I bet there is one which fits.

 

[pellman]

Thanks, fresh_42