Expanding a function with a power series

Question

How would I expand the following function as a power series, around $\eta=0$?

$$g_0(1,\eta)=\frac{\left(\frac{PV}{NkT}\right)_0-1}{4\eta}$$

Note that:

$$\left(\frac{PV}{NkT}\right)_0=1+\frac{3\eta}{\eta_c-\eta}+\sum_{k=1}^4kA_k\left(\frac{\eta}{\eta_c}\right)^k$$

Then we have:

$$g_0(1,\eta)=\frac{\frac{3\eta}{\eta_c-\eta}+\sum_{k=1}^4kA_k\left(\frac{\eta}{\eta_c}\right)^k}{4\eta}$$

Answer 1

Starting from your last equation: $$g_0(1,\eta)=\frac{\frac{3\eta}{\eta_c-\eta}+\sum_{k=1}^4kA_k\left(\frac{\eta}{\eta_c}\right)^k}{4\eta}$$ you have: $$g_0(1,\eta)=\frac{3}{4(\eta_c-\eta)}+\sum_{k=1}^4\frac k{4\eta_c^k}A_k\eta^{k-1}$$ The terms after the + sign are ok, you just need to deal with the expansion of $(\eta_c-\eta)^{-1}$. You can give $\eta_c$ as a factor, then you have $$\frac3{3\eta_c}\left(1-\frac{\eta}{\eta_c}\right)^{-1}$$ For $|x|<1$ you have $$\frac 1{1-x}=x+x^2+x^3+...=\sum_{n=1}^\infty x^n$$

Answer 2

Hint:

The sum is a polynomial with no constant term and its handling is not a problem.

Then

$$\frac{\eta}{\eta_c-\eta}=1-\frac1{1-\dfrac\eta{\eta_c}}=\frac\eta{\eta_c}+\frac{\eta^2}{\eta_c^2}+\frac{\eta^3}{\eta_c^3}+\cdots$$