Sum of a Series that is relevant to Fourier Transform

Question

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 $0<b<1, \ b\in {\rm I\!R}$, evaluate $f(t)$.

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 1

$$f(t)=\sum_{n=0}^{\infty}{b^n\,\sin(nt)}$$ 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

Answer 2

Since $b$ is real, it's probably easier to consider the series as the imaginary part of $$\sum_{n=1}^{\infty} b^n e^{int} = \sum_{n=1}^{\infty} (be^{it})^n = \frac{be^{it}}{1-be^{it}}. $$ One can then multiply by the conjugate to find $$ \frac{be^{it}}{1-be^{it}} \frac{1-be^{-it}}{1-be^{-it}} = \frac{be^{it}-b^2}{1+b^2-2b\cos{t}}, $$ so the imaginary part is $$ \frac{b\sin{t}}{1+b^2-2b\cos{t}}. $$

($b^{-n}$ is rather alarming to see in your calculation, since it doesn't converge to zero, so I suspect there may be something incorrect in that expression: it should probably be $\sum_{n \neq 0} b^{\lvert n \rvert}e^{int}/2i$. )