So there is an example in my book where g(k) is converted to G(f) and its written
$$g(k)\Longleftrightarrow G(f)$$
So:
$$a^ku(k)\Longleftrightarrow \frac{1}{1-ae^{-j2\pi f}}$$
My question is, how do we find that $a^ku(k)\Longleftrightarrow 1/(1-ae^{-j2 \pi f})$ ?
From the definition of the Fourier transform for series you have:
$$G(f)=\sum_{k=-\infty}^{\infty}g(k)e^{-j2\pi kf}=\sum_{k=0}^{\infty}a^ke^{-j2\pi kf}=\frac{1}{1-ae^{-j2\pi f}},\quad |a|<1$$
where the last equality follows from the formula of the geometric sum, which is valied if $|a|<1$ holds.
