Does the DTFT of $\frac{u[n-1]}{n}$ exist?
Note that the sequence

$$x[n]=\frac{u[n-1]}{n}\tag{1}$$

is in $\ell^2(\mathbb Z)$ because

$$\sum_{n\in\mathbb{Z}}|x[n]|^2=\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}&lt;\infty\tag{2}$$

but it is not in $\ell^1(\mathbb Z)$ since

$$\sum_{n\in\mathbb{Z}}|x[n]|=\sum_{n=1}^{\infty}\frac{1}{n}=\infty\tag{3}$$

We know that for sequences in $\ell^1(\mathbb{Z})$ the DTFT $X(e^{j\omega})$ exists and it is a continuous function of $\omega$. However, it is not necessary for a sequence to be in $\ell^1(\mathbb{Z})$ for the DTFT to exist. We can extend the definition of the DTFT to sequences in $\ell^2(\mathbb{Z})$, which implies that the DTFT converges in a mean square sense (and not uniformly), and that $X(e^{j\omega})$ may have discontinuities.

In that extended sense, the DTFT of $(1)$ exists, just as it exists for the ideal discrete-time differentiator or the discrete-time Hilbert transformer, the impulse responses of which also only decay as $1/n$.

The DTFT of $x[n]$ is

$$X(e^{j\omega})=\sum_{n=1}^{\infty}\frac{e^{-jn\omega}}{n}\tag{4}$$

The series $(4)$ is the complex Mercator series

$$\sum_{n=1}^{\infty}\frac{z^{n}}{n}=-\log(1-z),\qquad |z|\le 1,\quad z\neq 1\tag{5}$$

with $z=e^{-j\omega}$. It is the Taylor series for $-\log(1-z)$ and it converges for all $|z|\le 1$, $z\neq 1$. From $(5)$ we can immediately write down the $\mathcal{Z}$-transform of $x[n]$:

$$X(z)=\sum_{n=1}^{\infty}\frac{z^{-n}}{n}=-\log(1-z^{-1}),\qquad |z|\ge 1,\quad z\neq 1\tag{6}$$

Since $(6)$ also converges for $|z|=1$ except for $z=1$, we can obtain the DTFT directly from $(6)$ by setting $z=e^{j\omega}$. For $e^{j\omega}=1$, i.e., $\omega=0$, the DTFT has a singularity.


  $$X(e^{j\omega})=-\log(1-e^{-j\omega})\tag{7}$$


Equation $(7)$ agrees with WolframAlpha's result.
Also take a look at this related answer.



Let's visually check if it is plausible that the series $(4)$ really converges to $(7)$. I've computed the sum over the first $100$ terms of $(4)$, and I've plotted the real and imaginary parts of that partial sum together with the real and imaginary parts of $(7)$:


From the figure, it appears that the series $(4)$ indeed converges to $(7)$ in a mean square sense. This is of course no proof but a simple sanity check.