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* Page found: Das ideale Bosegas (eq math.2556.5)

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\begin{align}
  & \left\langle {{N}_{j}} \right\rangle =\sum\limits_{{{N}_{j}}=0}^{\infty }{{}}{{N}_{j}}p({{N}_{j}})=\sum\limits_{{{N}_{j}}=0}^{\infty }{{}}{{N}_{j}}\left( 1-{{t}_{j}} \right){{t}_{j}}^{{{N}_{j}}}=\left( 1-{{t}_{j}} \right){{t}_{j}}\frac{d}{d{{t}_{j}}}\sum\limits_{{{N}_{j}}=0}^{\infty }{{}}{{t}_{j}}^{{{N}_{j}}} \\ 
 & =\left( 1-{{t}_{j}} \right){{t}_{j}}\frac{d}{d{{t}_{j}}}\left( \frac{1}{1-{{t}_{j}}} \right)=\left( 1-{{t}_{j}} \right){{t}_{j}}\left( \frac{1}{{{\left( 1-{{t}_{j}} \right)}^{2}}} \right)=\frac{{{t}_{j}}}{\left( 1-{{t}_{j}} \right)} \\ 
\end{align}

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⟨ N j ⟩ = ∑ N j = 0 ∞ N j ⁢ p ⁢ ( N j ) = ∑ N j = 0 ∞ N j ⁢ ( 1 - t j ) ⁢ t j N j = ( 1 - t j ) ⁢ t j ⁢ d d ⁢ t j ⁢ ∑ N j = 0 ∞ t j N j = ( 1 - t j ) ⁢ t j ⁢ d d ⁢ t j ⁢ ( 1 1 - t j ) = ( 1 - t j ) ⁢ t j ⁢ ( 1 ( 1 - t j ) 2 ) = t j ( 1 - t j ) missing-subexpression delimited-⟨⟩ subscript 𝑁 𝑗 superscript subscript subscript 𝑁 𝑗 0 subscript 𝑁 𝑗 𝑝 subscript 𝑁 𝑗 superscript subscript subscript 𝑁 𝑗 0 subscript 𝑁 𝑗 1 subscript 𝑡 𝑗 superscript subscript 𝑡 𝑗 subscript 𝑁 𝑗 1 subscript 𝑡 𝑗 subscript 𝑡 𝑗 𝑑 𝑑 subscript 𝑡 𝑗 superscript subscript subscript 𝑁 𝑗 0 superscript subscript 𝑡 𝑗 subscript 𝑁 𝑗 missing-subexpression absent 1 subscript 𝑡 𝑗 subscript 𝑡 𝑗 𝑑 𝑑 subscript 𝑡 𝑗 1 1 subscript 𝑡 𝑗 1 subscript 𝑡 𝑗 subscript 𝑡 𝑗 1 superscript 1 subscript 𝑡 𝑗 2 subscript 𝑡 𝑗 1 subscript 𝑡 𝑗 {\displaystyle{\displaystyle\begin{aligned} &\displaystyle\left\langle{{N}_{j}% }\right\rangle=\sum\limits_{{{N}_{j}}=0}^{\infty}{{}}{{N}_{j}}p({{N}_{j}})=% \sum\limits_{{{N}_{j}}=0}^{\infty}{{}}{{N}_{j}}\left(1-{{t}_{j}}\right){{t}_{j% }}^{{{N}_{j}}}=\left(1-{{t}_{j}}\right){{t}_{j}}\frac{d}{d{{t}_{j}}}\sum% \limits_{{{N}_{j}}=0}^{\infty}{{}}{{t}_{j}}^{{{N}_{j}}}\\ &\displaystyle=\left(1-{{t}_{j}}\right){{t}_{j}}\frac{d}{d{{t}_{j}}}\left(% \frac{1}{1-{{t}_{j}}}\right)=\left(1-{{t}_{j}}\right){{t}_{j}}\left(\frac{1}{{% {\left(1-{{t}_{j}}\right)}^{2}}}\right)=\frac{{{t}_{j}}}{\left(1-{{t}_{j}}% \right)}\\ \end{aligned}}}

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⟨Nj⟩=∑Nj=0∞Njp(Nj)=∑Nj=0∞Nj(1−tj)tjNj=(1−tj)tjddtj∑Nj=0∞tjNj=(1−tj)tjddtj(11−tj)=(1−tj)tj(1(1−tj)2)=tj(1−tj)

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