How to turn partition sum into an integral?

[LightPhoton]

Homework Statement

How to evaluate partition sum in the limit where $$kT\gg\epsilon$$

Relevant Equations

$$Z_{tot}=\sum_0^\infty (2j+1)e^{-j(j+1)\epsilon/kT}$$

In, *An Introduction to Thermal Physics, page 235*, Schroder wants to evaluate the partition function

$$Z_{tot}=\sum_0^\infty (2j+1)e^{-j(j+1)\epsilon/kT}$$

in the limit that $kT\gg\epsilon$, thus he writes

$$Z_{tot}\approx\int_0^\infty (2j+1)e^{-j(j+1)\epsilon/kT}\,dj$$

But how is this correct? There was no factor of $j$ in the sum that could be replaced with $dj$. Also, it is good that $j$ is just a number, otherwise even the dimensions of $Z_{tot}$ would be wrong.

 

[renormalize]

There was no factor of $j$ in the sum that could be replaced with $dj$.

Yes, there is such a factor if you express "$1$" in terms of the summation index "$j$":$$Z_{tot}=\sum_{0}^{\infty}(2j+1)e^{-j(j+1)\epsilon/kT}\times1=\sum_{0}^{\infty}(2j+1)e^{-j(j+1)\epsilon/kT}\left((j+1)-j\right)\equiv\sum_{0}^{\infty}(2j+1)e^{-j(j+1)\epsilon/kT}\triangle j$$Now can you see how the integral approximates the sum?