Stirling's approximation for the entropy [closed]

I am trying to derive eq. (2.15) on page 34 of Birger Bergersen's and Michael Plischke's textbook on Equilibrium Statistical Mechanics second edition.

We have equation $S(E,V,N)=k_B \log \Omega(E,V,N)$. $$(2.14) \ \ \ \ \ \ \Omega(E,V,N) = \frac{V^N}{h^{3N}N!} \frac{(2\pi m E)^{3N/2}}{(3N/2-1)!}\frac{\delta E}{E}$$

Now, they say that, if $|\log(\delta E/E)|\ll N$, we find, using Stirling's approximation, $\log N! \approx N\log N-N$,

$$S(E,V,N) \approx Nk_B\log V/N + \frac{3N}{2}k_B\log(\frac{4\pi mE}{3Nh^2})+5Nk_B/2$$

Now, I did the calculation, but I get something else:

$$S(E,V,N)\approx k_B(N\ln V - N\ln N+N)+k_B\bigg[ \frac{3N}{2}\ln\frac{2\pi m E}{h^2}-(3N/2-1)\ln(3N/2-1)-(3N/2-1) \bigg]$$

But how to continue from here, I assume because $N\gg 1$ that $\ln(3N/2-1)\approx \ln(3N/2)$,and $3N/2\gg 1$ but then I get a term with $-Nk_B/2$ and not as it's written in the textbook with $5Nk_B/2$.

Where did I go wrong here?

Edit: I forgot to say that I neglected $\ln (\delta E/E)$, since it's much less than $N$.

closed as off-topic by Kyle Kanos, stafusa, JamalS, Jon Custer, JMacOct 10 '17 at 18:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Kyle Kanos, stafusa, JamalS, Jon Custer, JMac
If this question can be reworded to fit the rules in the help center, please edit the question.

Firstly, I think you have a typo in the expression you derived. The extreme last term, $\left(\frac{3N}{2}-1\right)$ should have a positive sign, and not negative. Other than that, you are on the right track.
You should get, $$S(E,V,N)\approx k_B(N\ln V-N\ln N +N)+ k_B\left[\frac{3N}{2}\ln\frac{2\pi mE}{\hbar^2}-\left(\frac{3N}{2}-1\right)\ln\left(\frac{3N}{2}-1\right)+\left(\frac{3N}{2}-1\right)\right]$$ You can immediately see that, $$k_B(N\ln V-N\ln N +N)=k_B\ln\frac{V}{N}+Nk_B$$ The term in square brackets can be approximated as, $$k_B\left[\frac{3N}{2}\ln\frac{2\pi mE}{\hbar^2}-\frac{3N}{2}\ln\frac{3N}{2}+\frac{3N}{2}\right]=\frac{3Nk_B}{2}\ln\frac{4\pi mE}{3N\hbar^2}+\frac{3Nk_B}{2}$$ Adding everything up, you get the required expression mentioned in the book.