Lagrange multipliers for Bose-Einstein distribution > We have $$ W = \prod_{s=1}^N \frac{(g_s-1+n_s)!}{(g_s-1)!n_s!} $$ For $ g_s,n_s>>1 $. We also have the constraints $$ \sum_{s=1}^Nn_s = n \hspace{10mm} \sum_{s=1}^Nn_sE_s = E $$ We are to show $ \ln{W}$ is greatest for $$ \boxed{n_s = g_s\left[e^{\beta(E_s-\mu)}-1\right]^{-1}}$$ Where $\beta$ and $\beta\mu$ are Lagrange multipliers. I have managed to reduce $\ln{W}$, using Stirling's approximation of $\ln{n!}=n\ln{n} - n$, to a horrible looking: $$ \ln{W} = \sum_{s=1}^N\left[(g_s-1+n_s)\ln{(g_s-1+n_s)} -(g_s-1)\ln{(g_s-1) -n_s\ln{n_s}}\right]$$ However, trying to use Lagrange multipliers gives an even worse expression. Help is much appreciated.

As $g_s,n_s>>1$, we have $$W = \prod_{s=1}^N \frac{(g_s-1+n_s)!}{(g_s-1)!n_s!}\approx\prod_{s=1}^N \frac{(g_s+n_s)!}{g_s!n_s!}$$ Using Stirling's approximation will then result in $$\ln{W} = \sum_{s=1}^N\left[(g_s+n_s)\ln{(g_s+n_s)} -g_s\ln{(g_s-1) -n_s\ln{n_s}}\right]$$ Consider the Lagrangian (where $\beta$ and $\lambda$ are the two Lagrange multipliers) $$\mathcal{L}=\ln{W}+\beta(E-\sum_{s=1}^Nn_sE_s)+\lambda(n-\sum_{s=1}^Nn_s)$$ Differentiating $\mathcal{L}$ with respect to $n_s$ and setting this to $0$ results in $$\begin{align}\frac{\partial\mathcal{L}}{\partial n_s}=\ln(g_s+n_s)-\ln n_s-\beta E_s-\lambda&=0\\\\\Rightarrow\ln\left(\frac{g_s+n_s}{n_s}\right)&=\beta E_s+\lambda\\\\\Rightarrow n_s&=g_s[e^{\beta E_s+\lambda}-1]^{-1}\end{align}$$ Without loss of generality, we can set $\lambda$ to $-\mu\beta$ leading to $$n_s=g_s[e^{\beta (E_s-\mu)}-1]^{-1}$$