Evaluating regular Fourier Transform from DFT
Your question makes perfect sense, yet your solution will not work. Note that the expression for $x[n]$

$$\begin{align}
x[n]=\frac{1}{N}\sum\limits_{k=0}^{N-1}X[k]e^{j\frac{2\pi}{N}kn} \end{align}$$

is periodic with period $N$. This means that the sum

$$
\frac{1}{N}\sum\limits_{n = -\infty}^{+\infty} e^{-j\Omega n}\sum\limits_{k=0}^{N-1}X[k]e^{j\frac{2\pi}{N}kn}$$

does not converge. Let's assume that $x[n]$ is defined for $0\le n&lt; N$ and is zero otherwise. Then we get for the discrete-time Fourier transform of $x[n]$

$$X(e^{j\Omega})=\sum_{n=0}^{N-1}x[n]e^{-jn\Omega}=
\frac{1}{N}\sum_{n=0}^{N-1}\sum\limits_{k=0}^{N-1}X[k]e^{j\frac{2\pi}{N}kn}e^{-jn\Omega}\tag{1}$$

This formula actually works and can be used if only the DFT $X[k]$ is given. Of course this only works for finite length signals $x[n]$. Equation (1) simply means that the DFT contains the complete information about the (finite-length) signal $x[n]$ and any frequency point $\Omega$ of the DTFT can be evaluated using the discrete values $X[k]$.