Sum of a Series that is relevant to Fourier Transform

I'm currently studying for a test that envelops both Fourier transforms and Fourier Series. This exercise is meant to build knowledge and give me a better understanding of the basics, however I think I might be missing some property of the sum of a series because I am stuck.

The problem is the following:

Be $f(t)=\sum_{n=0}^{\infty}{b^nsin(nt)}$ where $0

Okay, using euler's formula for $sin(nt)$ and manipulation of the summation variables I arrived at the following expression:

$$f(t)=\sum_{n=-\infty}^{\infty}{\frac{(b^n-b^{-n})}{2i}e^{int}}$$ and I feel that those b's look a lot like some kind of hyperbolic euler identity, but I am unsure whether that is a good road to go to, or wheter I should have went another way entirely from the beggining.

The answer at the end states that $f(t)=\frac{bsin(t)}{1-2bcos(t)+b^2}$, if that helps anyone.

Answer

$$f(t)=\sum_{n=0}^{\infty}{b^n\,\sin(nt)}$$
Converting to exponential:
$$b^n\,\sin(nt)=\frac{1}{2} i b^n \left(e^{-i n t}-e^{i n t}\right)=\frac{1}{2} i\left[(be^{-it})^n-(be^{it})^n \right]=\ldots$$
Calculating the sums we have
$$\sum _{n=0}^{\infty } \left(b e^{-i t}\right)^n=\frac{1}{1-b e^{-i t}};\;\sum _{n=0}^{\infty } \left(b e^{i t}\right)^n=\frac{1}{1-b e^{i t}}$$
and then
$$\ldots\ =\ \frac{1}{2} i\left(\frac{1}{1-b e^{-i t}}- \frac{1}{1-b e^{i t}}\right)\ =\ \frac{1}{2} i\,\frac{1-b e^{i t}-1+b e^{-i t}}{(1-b e^{-i t})(1-b e^{i t})}=\ldots$$
$$\ldots=\frac{1}{2i} \,\frac{b e^{i t}-b e^{-i t}}{(1-b e^{-i t})(1-b e^{i t})}=\frac{b\sin t}{b^2-b \left(e^{i t}+e^{-i t}\right)+1}=\frac{b\sin t}{b^2-2b\cos t+1}$$

Hope this helps