What is the z-transform of $\frac{1}{n}$

Question

given that $n \geq 1$ and $x(n) = \frac{1}{n}$, how can I get the z-transform of $x(n)$? Thanks

Answer 1

I do not know too much, but:

In this Book on Table A.6.5 number 32 the result is correct.

Answer 2

We are looking for $$X(z)=\sum_{n=1}^{\infty}\frac{z^{-n}}{n}$$

Consider the following Taylor series expansion: $$\log(1-y)=-\sum_{n=1}^{\infty}\frac{y^{n}}{n}, \;\text{for }|y|<1$$ The desired transform can be calculated assuming $y=z^{-1}$:

$$X(z)=-\log(1-z^{-1})=\log\left(\frac{1}{1-z^{-1}}\right)=\log\left(\frac{z}{z-1}\right),\;\text{for }|z|>1$$