Why does a longer observation time improve DFT resolution, but repeating a signal does not? As was proven here: https://math.stackexchange.com/questions/228614/why-doesnt-repeating-a-signal-give-rise-to-a-finer-resolution-of-dft-fft repeating a certain sequence does not improve DFT frequency resolution.
However, it is known that oberserving a signal for a longer time will in fact improve the ability to resolve two tones in a signal (because the 'beamwidth' of the window in the frequency domain will become narrower: it is 2/NT with NT the total observation time).
I do not understand this discrepancy: If we assume the signal has two tones f1 and f2, and has a period T. Let's say we observe it over a time N*T such that 1/(NT) becomes small enough to distinguish f1 and f2. We have improved the frequency resolution by observing the signal over a longer time. Why is this not equivalent to simply observing the signal for 1 period, and then paste it together N times? This will give the same signal in the time domain, but according to the proof in the link that I posted, it will not improve the frequency resolution.
The only explanation that I can think of myself is: observing for a longer time period does not improve the frequency resolution further once the observation time becomes larger than T. However, I have written some matlab code where I do an FFT on a pure sine over a certain time. I clearly see the 'beam width' in the frequency domain associated with the windowing start to narrow down as I increase the observation time. Given that a pure sine is also periodic and the observation times that I used are bigger than this period, I must thus conclude that my 'explanation' is incorrect.
Could someone try to show me the correct explanation?
Why is this not equivalent to simply observing the signal for 1 period, and then paste it together N times?

It's only equivalent if certain conditions are met. Let's look at a single sine wave with frequency $f_0$ and period $T_0 = 1/f_0$. Sample rate is $f_s$, sampling period $T_s = 1/f_s$ and DFT length is $N$.
The input to the FFT would be
$$x[n] = sin(2\pi \cdot f_0\cdot T_s \cdot n), 0 \leq n &lt; N$$
Now let's create two signals of length $2N$: one with double the length and one with pasting two buffers together.
$$x_1[n] = sin(2\pi \cdot f_0\cdot T_s \cdot n), 0 \leq n &lt; 2N$$
and
$$x_1[n] = sin(2\pi \cdot f_0\cdot T_s \cdot mod(n,N)), 0 \leq n &lt; 2N$$
The only difference is that modulo function in the second one, but it's important. It basically resets the counter to 0 when it reaches an integer multiple of N. In general this will create a discontinuity in the signal. It forces $x_2[N] = x_2[0] = 0$ even if $x[N]$ would be non zero. The first halves of $x_1$  and $x_2$ are the same but the second halves are not. $x_1$ is still a sine wave but $x_2$ isn't anymore.
The ONLY case where pasting is the same as making the signal longer is if the period of the signal is an integer multiple of the sampling period. In this case, there is no discontinuity and both cases are identical. And sure enough, in this case the DFT will resolve two different frequencies perfectly (as long as both meet the integer multiple condition).
A more philosophical interpretation: Making the signal longer provides more information. Pasting buffers just repeats information you already have, so you are not gaining anything new.