NMR: How the relaxation times T1 and T2 depend on the correlation time / amount of molecular tumbling.
Relaxation is often very counterintuitive and isn't a particularly easy subject to understand, so it's not always useful to argue based on what seems to be logical. I'm not sure where you're studying this from, but it seems to be slightly lacking in the detail needed to properly understand what's going on. Here is a brief overview of what's going on, but I strongly recommend consulting a textbook which will explain it in more detail. Chapter 9 of James Keeler's Understanding NMR Spectroscopy (2nd ed.) is a very good reference. You can find a pdf of the corresponding chapter in the previous edition here.

The link between rate of tumbling (correlation time, $\tau_c$) and the "amount of fluctuations at the right frequency" is given by the spectral density

$$j(\omega) = \frac{2\tau_c}{1 + \omega^2\tau_c^2}$$

This function tells you how much the random fields are fluctuating at each frequency $\omega$. For relaxation, the important frequency to look at is the Larmor frequency $\omega_0$, i.e. the central quantity is

$$j(\omega_0) = \frac{2\tau_c}{1 + \omega_0^2\tau_c^2}$$

By differentiating this with respect to $\tau_c$ you can find that this is at a maximum when $\tau_c = 1/\omega_0$. The rate of spin-lattice (longitudinal) relaxation is proportional to $j(\omega_0)$, so it is fastest when $\tau_c = 1/\omega_0$. That means that $T_1$ will be smallest when $\tau_c = 1/\omega_0$. When $\tau_c$ decreases below this, or increases above this, spin-lattice relaxation will be slowed down and $T_1$ will increase.

On the other hand, spin-spin (transverse) relaxation depends on the sum $j(\omega_0) + j(0)$. $j(0)$ is simply equal to $2\tau_c$, and you can show (again with calculus) as $\tau_c$ increases, the total sum $j(\omega_0) + j(0)$ increases continually. So $T_2$ decreases when $\tau_c$ increases.