How is the Series ε + 2 ε^2 + 3 ε^3 + … Equal to ε/(1- ε)^2?

[jbowers9]

What they said:

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?

 

[NoMoreExams]

I'm not sure I understand your notation, please use proper parantheses.

Do you mean:

f(ε) = Σ e-nε/kT n = 0 to ∞

$$f(\epsilon) = \sum_{n=0}^{\infty} e^{-\frac{n \epsilon}{kT}$$

OR

$$f(\epsilon) = \sum_{n=0}^{\infty} \frac{e^{-n \epsilon}}{kT}$$

And what is your question?

Is it why does this hold:

$$\epsilon + 2 \epsilon^{2} + 3 \epsilon^{3} = \frac{\epsilon}{\left(1 - \epsilon \right)^2}$$

If so, that's just taking derivative of both sides. If you need more clarification, let me know

 

[Mute]

I'm not sure I understand your notation, please use proper parantheses.

Do you mean:

f(ε) = Σ e-nε/kT n = 0 to ∞

$$f(\epsilon) = \sum_{n=0}^{\infty} e^{-\frac{n \epsilon}{kT}$$

OR

$$f(\epsilon) = \sum_{n=0}^{\infty} \frac{e^{-n \epsilon}}{kT}$$

And what is your question?

Is it why does this hold:

$$\epsilon + 2 \epsilon^{2} + 3 \epsilon^{3} = \frac{\epsilon}{\left(1 - \epsilon \right)^2}$$

If so, that's just taking derivative of both sides. If you need more clarification, let me know

The sum is the first one. The factor kT pegs this as a statistical mechanices problem.

Basically, to the OP, what the problem amounts to is that you have an infinite sum,

$$\sum_{n=0}^{\infty}\exp\left[-\frac{n\varepsilon}{k_BT}\right]$$

which is really just a geometric series: if you let $x = \exp<br /> \left[-\frac{\varepsilon}{k_BT}\right]$, you get

$$\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$$

(since the exponential is always less than one). Hence, if you take a derivative with respect to $\beta = 1/(k_BT)$ you get

$$\sum_{n=0}^{\infty} n \varepsilon e^{-n \varepsilon \beta} = -\frac{\varepsilon e^{-\varepsilon \beta}}{(1-e^{-\varepsilon \beta})^2}$$

which allows you to easily find the average energy of the system.

Or, even simpler, leaving it as x and differentiating with respect to that,

$$\sum_{n=0}^{\infty} nx^{n-1} = \frac{1}{(1-x)^2}$$

so just multiply by $\varepsilon e^{-\varepsilon \beta}$ to get

$$\sum_{n=0}^{\infty} n \varepsilon x^n = \frac{\varepsilon x}{(1-x)^2}$$

 

[jbowers9]

And what is your question?

Is it why does this hold:

ε + 2 ε2 + 3 ε3 + … = ε/(1- ε)2


If so, that's just taking derivative of both sides. If you need more clarification, let me know

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

and what does it mean?

I understand how to proceed to get the expression

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

but epsilon is a constant, hυ, not e-nε/kT. Do they mean that ε = f(ε) = e-nε/kT? That makes sense then.