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.


Note that the sequence


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


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


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


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.


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 $(7)$: enter image description here

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.

  • $\begingroup$ WolframAlpha thinks!?!?! We've reached the singularity, then! :D $\endgroup$
    – Peter K.
    Jan 3 '18 at 16:43
  • 1
    $\begingroup$ @PeterK.: yeah, it's when those funny dots appear, then it's thinking hard ... :) $\endgroup$
    – Matt L.
    Jan 3 '18 at 16:47
  • 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 $\endgroup$
    – Juancho
    Jan 4 '18 at 1:35
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    $\begingroup$ @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 '18 at 16:30

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