$\sum_{m_{3}} e^{- β u s n} = e^{0} + e^{- β u s} + e^{- β 2 u s} + \dots$
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$A = \frac{1}{1 - e^{- β u s}}$

Answer

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Answer

$A = \frac{1}{1 - e^{- β u s}}$

Steps

Step 1 ：Given summation: $\sum_{m_{3}} e^{- β u s n} = e^{0} + e^{- β u s} + e^{- β 2 u s} + \dots$

Step 2 ：Recognize this as a geometric series with a common ratio of $e^{- β u s}$

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

Step 4 ： $A = \frac{1}{1 - e^{- β u s}}$