What is the Maclaurin Series for f(x)=x/(e^x-1) up to x^2?

DjDriftX
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Find the first few terms of the Taylor Series around x=0...?
of the function

f(x)= {x/(e^x - 1) , x =/ 0}
{1 , x=0}

the function is piecewise.
up to and including the term involving x^2

It says to not compute derivatives of f but to use the formula for the Taylor series of e^x


x/(e^x-1) = x (1/(e^x-1)
so.
would that be x (1/ \sum(xn/n!) - 1)
or maybe x \sum (1 / (xn/n!) - 1))

I'm not really sure where to start
 
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Remember you only need terms up to and including x^2. 1-e^x=x+x^2/2!+x^3/3!+... Divide numerator and denominator by x. Now you've got 1/(1+x/2!+x^2/3!+...). You know an expansion for 1/(1+a), right? Use it. Throw away terms that are higher power than you are looking for.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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