Given
The ratio of number of oscillators in their (n+1)th quantum state of excitation to the number in their nth quantum state is $$N_n+1/N_n = \exp(-\hbar\omega/K_bT)$$
Thus the fraction of the total number of oscillators in the nth quantum state is $$ \frac{N_n}{\sum_{s=0}^\infty N_s}=\frac{\exp(-n\hbar\omega/K_bT)}{\sum_{s=0}^\infty \exp(-s\hbar\omega/K_bT)}$$ Can someone explain the math behind how to arrive at equation 2 from equation 1
Assuming the ratio is in fact $N_{n+1}/N_n$, which completely makes sense given OP's first sentence, we can write $\exp(-\hbar\omega/K_bT) = c$ and we know that $N_{n+1}/N_n = c \implies N_n=c^nN_0$. Thus we trivially obtain: $$ \require{cancel} \frac{N_n}{\sum_{s=0}^\infty N_s} =\frac{c^{n}\cancel{N_0}}{\sum_{s=0}^\infty c^{s}\cancel{N_0}} =\frac{c^{n}}{\sum_{s=0}^\infty c^{s}} =\frac{\exp(-\hbar\omega/K_bT)^{n}}{\sum_{s=0}^\infty \exp(-\hbar\omega/K_bT)^{s}} =\frac{\exp(-n\hbar\omega/K_bT)}{\sum_{s=0}^\infty \exp(-s\hbar\omega/K_bT)} $$