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f(ε) = Σ e-nε/kT n = 0 to ∞

Write the following series in terms of f’(ε):

g(ε) = Σ n ε e-nε/kT n = 0 to ∞

Then use the geometric series results to show that g can be written in the form:

ε + 2 ε2 + 3 ε3 + … = ε/(1- ε)2

What I did:

Let x = e-nε/kT

1/(1- x) = 1 + x + x2 + ...

1/(1- x)2 = 1 + 2x + 3x2 + 4x3 + ...

x/(1- x)2 = x + 2x2 + 3x3 + 4x4 +...

which leads me to:

g(ε) = ε e-nε/kT / (1 - e-nε/kT)2 n = 0 to ∞

How do they get, ε + 2 ε2 + 3 ε3 + … = ε/(1- ε)2

and what does it mean?

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# F(ε) = Σ e-nε/kT n = 0 to ∞

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