I have to compute the following integral:
$$N = -\frac{2\pi A}{h^2}\int_{0}^{\infty} \ln(1-ze^{-\beta pc})p\mathrm{d}p$$
Let's say $z << 1$. Thus:
$$\ln(1-x) = -\sum_{n=1}^{\infty} \frac{x^n}{n}$$
$$N = \frac{2\pi A}{h^2}\int_{0}^{\infty} \sum_{n=1}^{\infty} \mathrm{d}p\frac{z^n e^{-n \beta pc}}{n}p$$
Now it is convenient to use the following change of variables:
$$x = \beta pc$$
$$N = \frac{2\pi A}{(h \beta c)^2}\int_{0}^{\infty} \sum_{n=1}^{\infty} \mathrm{d}x\frac{z^n e^{-n x}}{n}x$$
It's time to factorize the series:
$$N = \frac{2\pi A}{(h \beta c)^2}\sum_{n=1}^{\infty} \frac{z^n}{n} \int_{0}^{\infty} xe^{-x} \mathrm{d}x$$
But I saw in the provided solution that I am missing a factor of three on the series, as it is stated to be:
$$N = \frac{2\pi A}{(h \beta c)^2}\sum_{n=1}^{\infty} \frac{z^n}{n^3} \int_{0}^{\infty} xe^{-x} \mathrm{d}x$$
From where is this to the power of 3 coming?
The issue is that you went from $e^{-nx}$ to $e^{-x}$ without doing a proper change of variable. If $y=nx$ then the $x$ that you have next to the exponential becomes $y/n$ and $dx$ becomes $dy/n$. Therefore you have two extra $1/n$ factors