Why does the series Taylor expand as e^-nx?

[unscientific]

In this section, they derive the Sommerfeld formula.

In the first step it seems like they have expanded $\frac{1}{(1+e^x))^2}$. I'm not sure why does the series taylor expand as $e^{-nx}$?

Also how did they get from the 2nd to the 3rd step?

Simply by comparing terms we see they are different:

For 2nd step we get terms of $x^se^x(-1 + 2e^{-x} - 3e^{-2x} + ...)$.
For 3rd step we get terms of $e^{-x} - 2e^{-2x} + 3e^{-3x} - ...$

 

[chingel]

Yeah in the second step it needs to be multiplied by -1 and the e^x should be e^(-x) instead.

To get the expansion, you could write \frac{1}{(1+e^x)^2} as \frac{e^{-2x}}{(1+e^{-x})^2}. Then you could do the binomial series expansion on it, or you could write it out using the geometric series formula, which gives \frac{e^{-2x}}{(1+e^{-x})^2}=e^{-2x}(1-e^{-x}+e^{-2x}-e^{-3x}+...)^2. Collecting the terms that come from squaring it, you get the appropriate formula.

 

[unscientific]

... \frac{e^{-2x}}{(1+e^{-x})^2}=e^{-2x}(1-e^{-x}+e^{-2x}-e^{-3x}+...)^2. Collecting the terms that come from squaring it, you get the appropriate formula.

Collecting the terms give the right terms, but how did they obtain a general expression $\sum_{n=0}^{\infty} e^{-x} [(n+1)(-1)^{n+1} e^{-nx}]$?

 

[Bill_K]

To expand $\frac{1}{(1 + e^{-x})^2}$, do it in terms of $y = e^{-x}$.
\frac{1}{1 + y} = \sum{(-1)^n y^n}
Differentiate both sides wrt $y$:
- \frac{1}{(1 + y)^2} = \sum{(-1)^n n y^{n-1}} = \sum{(-1)^{n+1} (n+1) y^n}

 

[unscientific]

To expand $\frac{1}{(1 + e^{-x})^2}$, do it in terms of $y = e^{-x}$.
\frac{1}{1 + y} = \sum{(-1)^n y^n}
Differentiate both sides wrt $y$:
- \frac{1}{(1 + y)^2} = \sum{(-1)^n n y^{n-1}} = \sum{(-1)^{n+1} (n+1) y^n}

Actually I think it's $e^x$ instead:

Starting, LHS is simply geometric series with factor $e^{-x}$:

\frac{1}{1+e^x} = \sum (-1)^n e^{-nx}

Differentiating both sides with respect to x,

-\frac{e^x}{(1+e^x)^2} = -\sum (-1)^{n+1} n e^{-nx}

\frac{e^x}{(1+e^x)^2} = \sum (-1)^{n+1} n e^{-nx}

This leads to the answer immediately.