Prove that $\frac{t^{x-1}}{1+t^y} = \sum_{n=0}^{\infty} (-1)^nt^{x-1+yn}$.

Question

Prove that $\frac{t^{x-1}}{1+t^y} = \sum_{n=0}^{\infty} (-1)^nt^{x-1+yn}$.

I tried expanding the exponential terms by substituting the definition of exp such that I have a series in the numerator and denominator but no matter how I tried manipulating the expressions, I wasn't able to show the equality.

Thanks in advance, whiterock.

Answer 1

$$\sum_{n=0}^{\infty} (-1)^nt^{x-1+yn} = \sum_{n=0}^{\infty} (-1)^nt^{x-1}(t^y)^n = t^{x-1}\sum_{n=0}^\infty(-t^y)^n = t^{x-1} \frac{1}{1-(-t^y)}=\frac{t^{x-1}}{1+t^y}$$ whenever the exponentials are well defined and $|t^y|<1$.