I have started learning DSP on my own and I have this doubt. I have done some googling but haven't found an answer. I hope that someone here would give the answer. It will be of great help.
1 Answer
Note that the sequence
$$x[n]=\frac{u[n-1]}{n}\tag{1}$$
is in $\ell^2(\mathbb Z)$ because
$$\sum_{n\in\mathbb{Z}}|x[n]|^2=\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}<\infty\tag{2}$$
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.
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$\begingroup$ WolframAlpha thinks!?!?! We've reached the singularity, then! :D $\endgroup$– Peter K. ♦Jan 3, 2018 at 16:43
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1$\begingroup$ @PeterK.: yeah, it's when those funny dots appear, then it's thinking hard ... :) $\endgroup$– Matt L.Jan 3, 2018 at 16:47
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2$\begingroup$ The question is about the existance of the FT. It does not converge since the region of convergence of the Z transform excludes the unit circle $\endgroup$– JuanchoJan 4, 2018 at 1:35
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1$\begingroup$ @Juancho: Thanks, I misread and worked out the Z-transform. However, note that the FT does exist as I'm going to show as soon as I have more time. $\endgroup$– Matt L.Jan 4, 2018 at 16:30