I would like to derive the inverse discrete Fourier transform from the Fourier series. This should be possible since the DFT is just the Fourier series of a discrete time signal but I have run into an infinite series that I cannot figure out how to eliminate. Help is appreciated.
I start with the Fourier series:
\begin{aligned} x(t) & =\sum_{k=-\infty}^{+\infty}C_{k}e^{i2\pi kt/T},\\ C_{k} & =\frac{1}{T}\int_{-T/2}^{T/2}x(t)e^{-i2\pi kt/T}\textrm{d}t. \end{aligned}
I sample $x(t)$ N times during its period $T$ and arrive at the DFT coefficients, using a Riemann sum approximation:
\begin{aligned} C_{k}&=\frac{1}{T}\int_{-T/2}^{T/2}x(t)e^{-i2\pi kt/T}\textrm{d}t\\&=\frac{1}{T}\sum_{n=0}^{N-1}x(nT/N)e^{-i2\pi kn/N}\left(\frac{T}{N}\right)\\&=\frac{1}{N}\sum_{n=0}^{N-1}x(nT/N)e^{-i2\pi kn/N}. \end{aligned}
Where I am stuck is the inverse transform:
\begin{aligned} x(t) & =\sum_{k=-\infty}^{+\infty}C_{k}e^{i2\pi kt/T}\\ x\left(\frac{nT}{N}\right) & =\sum_{k=-\infty}^{+\infty}C_{k}e^{i2\pi kn/N}\\ & =\sum_{h=-\infty}^{+\infty}\sum_{k=0}^{N-1}C_{k+hN}e^{i2\pi(k+hN)n/N}\\ & =\sum_{h=-\infty}^{+\infty}\sum_{k=0}^{N-1}C_{k+hN}e^{i2\pi kn/N}\\ & =\ldots ? \end{aligned}
Since the equation should be:
\begin{aligned} x\left(\frac{nT}{N}\right)&=\sum_{k=0}^{N-1}C_{k}e^{i2\pi kn/N} \end{aligned}
It appears there must be a way to eliminate all values where $h\neq 0$. All textbooks I've read say that because it is periodic, there is no need to look at other periods, but I do not see how to arrive at this analytically.
Can someone please point out the key piece I'm missing?