At the end of 3b1b's fourier series video, he poses a challenge to the viewer about a $step(t)$ function that starts at 1 from 0 to 0.5 and -1 from 0.5 to 1.
$C_n=\frac{2}{n\pi i}$ for odd n and zero otherwise.
The second challenge says that by using the complex definition of sine: $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ show that:
$step(t)=\sum_{n=-\infty}^{\infty} C_ne^{2\pi nit}=\sum_{n=1,3,5,...}^{} \frac{4}{n\pi}\sin(2\pi nt)$
I've got... close? Basically if you re-arrange the complex definition then: $2isin(2\pi nt)+e^{2\pi int}=e^{2\pi int}$
So substituting that into the equation you'd get:
$\frac{2}{n\pi i}(2i\sin(2\pi nt)+e^{2\pi int})$ which further simplifies to $\frac{4}{n\pi}\sin(2\pi int)+\frac{2}{n\pi i}e^{-2\pi ti}$ which seems very close to what it should be but I dont know how to get rid of the $\frac{2}{n\pi i}e^{-2\pi ti}$
Or maybe I'm just going down the wrong path and the answer is way more obvious.
Write the first sum separately for positive and negative numbers: $$step(t)=\sum_{n=-\infty}^{-1}C_ne^{2\pi nit}+C_0+\sum_{n=1}^{\infty}C_ne^{2\pi nit}$$ Now let's change the variable in the first sum from $n$ to $m=-n$ $$step(t)=\sum_{m=\infty}^{1}C_{-m}e^{2\pi (-m)it}+C_0+\sum_{n=1}^{\infty}C_ne^{2\pi nit}$$ Note that $C_{-m}=-C_m$, so $$step(t)=-\sum_{m=1}^\infty C_m e^{2\pi (-m)it}+C_0+\sum_{n=1}^{\infty}C_ne^{2\pi nit}$$ Now you can write this as a single sum $$step(t)=C_0+\sum_{n=1}^\infty C_n(e^{2\pi nit}-e^{2\pi(-n)it})$$ Now notice that all even $C_n$ are zero. Plug in the value of $C_n$ and write the difference of exponentials as a sine function.