What is the z-transform of $\frac{1}{n}$ given that $n \geq 1$ and $x(n) = \frac{1}{n}$, how can I get the z-transform of $x(n)$?
Thanks
 A: I do not know too much, but:

In this Book on Table A.6.5 number 32 the result is correct.
A: 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 $y=z^{-1}$:
$$X(z)=-\log(1-z^{-1})=\log\left(\frac{1}{1-z^{-1}}\right)=\log\left(\frac{z}{z-1}\right),\;\text{for }|z|>1$$
A: Good Question 
$$\frac{u(n-1)}{n} \rightleftharpoons  ??$$ 
We know that Z transform is defined as :
$$X(z) = \sum_{n=-\infty}^{\infty}x(n)z^{-n}$$
$$u(n) \rightleftharpoons \frac{z}{z-1}$$
By the use of time shifting property
$$x(n-1) \rightleftharpoons z^{-1} X(z)$$
$$u(n-1) \rightleftharpoons \frac{1}{z-1}$$
Differentiate the General formula of X(z) with respect to z
$$\frac{dX(z)}{dz} = -\sum_{n=-\infty}^{\infty}nx(n)z^{-n-1}$$
$$\frac{zdX(z)}{dz} = -\sum_{n=-\infty}^{\infty}nx(n)z^{-n}$$
$$nx(n) \rightleftharpoons -\frac{zdX(z)}{dz}$$
Now we want to find out the Z transform of $$\frac{u(n-1)}{n}$$
$$x(n) = \frac{u(n-1)}{n} \rightleftharpoons X(z)$$
$$n x(n) \rightleftharpoons -\frac{zdX(z)}{dz}$$
$$n [\frac{u(n-1)}{n}] \rightleftharpoons -\frac{zdX(z)}{dz}$$
$$u(n-1) \rightleftharpoons -\frac{zdX(z)}{dz}$$
Means
$$-\frac{zdX(z)}{dz} = \frac{1}{z-1}$$
$$\frac{dX(z)}{dz} = -\frac{1}{z(z-1)}$$
$$\frac{dX(z)}{dz} = [\frac{1}{z}] + [\frac{1}{z-1}]$$
$$X(z) = ln(\frac{z}{z-1})$$
Hence
$$\frac{u(n-1)}{n} \rightleftharpoons ln(\frac{z}{z-1})$$
Now we have some interesting series for natural log
$$\frac{1}{z} + \frac{1}{2z^2} + \frac{1}{3z^3} + ...... = ln(\frac{z}{z-1})$$
For z = 2
$$\frac{1}{2} + \frac{1}{2\times2^2} + \frac{1}{3\times2^3} + ...... = ln(2)$$
For  $$z = i^2 = -1$$
$$\frac{1}{1} - \frac{1}{2} + \frac{1}{3} + ...... = ln(2)$$
