7

Show that,

$$\sum_{n=1}^{\infty}{x^{n-1}\over(n-1)!}\cdot{e^{-xn}-1\over e^{xn}-1}=-e^{-x(1-e^{-x})}$$

My try:

We know

$$\sum_{n=1}^{\infty}{x^{n-1}\over (n-1)!}=e^x$$

$$\sum_{n=1}^{\infty}{e^{-xn}-1\over e^{xn}-1}={1\over 1-e^{x}}$$

How do I use these two formulae to arrive at the top formula?

Let and try letting $x=\ln{y}$ then we have

$$\sum_{n=1}^{\infty}{\ln{y}^{n-1}\over (n-1)!}\cdot{y^{-n}-1\over y^n-1}=-y^{{1\over y}-1}$$

Still can't see anything obvious step to do next, any hints?

Marco Cantarini
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gymbvghjkgkjkhgfkl
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1 Answers1

7

Note that $$\frac{e^{-xn}-1}{e^{xn}-1}=\frac{1}{e^{nx}}\frac{e^{-xn}-1}{1-e^{-nx}}=-e^{-nx} $$ hence $$\sum_{n\geq1}\frac{x^{n-1}}{\left(n-1\right)!}\frac{e^{-xn}-1}{e^{xn}-1}=-\sum_{n\geq1}\frac{x^{n-1}}{\left(n-1\right)!}e^{-xn} $$ $$=-e^{-x}\sum_{n\geq0}\frac{x^{n}}{n!}e^{-xn}=\color{red}{-\exp\left(-x+xe^{-x}\right)}.$$

Marco Cantarini
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