\[
\sum_{m_{3}} e^{-\beta u s n}=e^{0}+e^{-\beta u s}+e^{-\beta 2 u s}+\cdots
\]
Converge para
\[
A=\frac{1}{1-e^{-\beta u s}}
\]

Answer

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Answer

\(\boxed{A = \frac{1}{1 - e^{-\beta u s}}}\)

Steps

Step 1 ：Given summation: \(\sum_{m_{3}} e^{-\beta u s n}=e^{0}+e^{-\beta u s}+e^{-\beta 2 u s}+\cdots\)

Step 2 ：Recognize this as a geometric series with a common ratio of \(e^{-\beta u s}\)

Step 3 ：Calculate the sum of the geometric series: \(A = \frac{1}{1 - e^{-\beta u s}}\)

Step 4 ：\(\boxed{A = \frac{1}{1 - e^{-\beta u s}}}\)