Using result for sum of geometric series prove

Question

$\frac{1}{(\alpha+e^{-x^2})}$ = ...

using the sum of the geometric series $1 + z + z^{2}+...+z^{n}$

Any help on this would be very appreciated as I'm stuck as to where to start. As in the first fraction the denominator is not 1.

Thanks in advance!

Answer 1

Here is a start :

$1+z+z^2+...+z^{n-1}=\frac{1-z^{n}}{1-z}$ thus using $z=-\frac{e^{-x^2}}{\alpha}$ you can write $\sum_{k=0}^{n-1}{z^k}+\frac{z^n}{1-z}=\frac{1}{1-z}$ replace $z$ and change the indices of the sum to make it start at $1$ and end at $n$...

Here is the solution :

You will have $\sum_{k=0}^{n-1}{(-\frac{e^{-x^2}}{\alpha})^k}+\frac{(-\frac{e^{-x^2}}{\alpha})^n}{1-(-\frac{e^{-x^2}}{\alpha})}=\sum_{k=0}^{n-1}{(-1)^k\frac{e^{-kx^2}}{\alpha^k}}+(-1)^n\frac{e^{-nx^2}}{\alpha^n(\alpha+e^{-x^2})}=\frac{1}{\alpha+e^{-x^2}}$ then by changing the indices you have $\sum_{k=1}^{n}{(-1)^{(k-1)}\frac{e^{-(k-1)x^2}}{\alpha^{k-1}}}+(-1)^n\frac{e^{-nx^2}}{\alpha^n(\alpha+e^{-x^2})}=\frac{1}{\alpha+e^{-x^2}}$

Answer 2

Hint: Start from the high school identity $$1-u^n=(1-u)(1+u+\dots+u^{n-1}),$$ and rewrite it as $$\frac1{1-u}=1+u+\dots+u^{n-1}+\frac{u^n}{1-u}.$$