Z-transform of $nu_{n}$ and $u_{n}/n$ I am studying z transform and I couldn't get how to derive these two formulas
$$Z[ nu_{n}] =-z \frac{d}{dz} Z(u_{n}) $$ 
$$Z\left[ \frac{1}{n} u_{n}\right] =- \int_0^z z^{-1}Z(u_{n}) $$ 
These formula are somewhat similar to the corresponding ones of Laplace transforms which also involves derivatives and integrals. Any idea on how to prove it?
Thanks.
 A: Look, the first one is pretty straightforward to prove:
For a discrete time domain function $f(k)$, its Z-Transform is defined as:
$$\mathcal{Z}\{f(k)\}=\sum_{k=-\infty}^{+\infty}f(k)z^{-k}$$
We can differentiate on both sides:
\begin{align}
\frac{d}{dz}\mathcal{Z}\{f(k)\}&=\frac{d}{dz}\sum_{k=-\infty}^{+\infty}f(k)z^{-k}\\
&=\sum_{k=-\infty}^{+\infty}\frac{d}{dz}f(k)z^{-k}\\
&=\sum_{k=-\infty}^{+\infty}f(k)\frac{d}{dz}z^{-k}\\
&=\sum_{k=-\infty}^{+\infty}-kf(k)z^{-k-1}\\
&=-z^{-1}\sum_{k=-\infty}^{+\infty}kf(k)z^{-k}\\
&=-z^{-1}\mathcal{Z}\{kf(k)\}\\
\end{align}
So, finally: 
$$\mathcal{Z}\{kf(k)\}=-z\frac{d}{dz}\mathcal{Z}\{f(k)\}$$
The integral property, on the other hand, is a little harder to prove with such a direct method, but let's go:
\begin{align}
\int\mathcal{Z}\{f(k)\}dz&=\int\left[\sum_{k=-\infty}^{+\infty}f(k)z^{-k}\right]dz\\
&=\sum_{k=-\infty}^{+\infty}\int f(k)z^{-k}dz\\
&=\sum_{k=-\infty}^{+\infty}f(k)\int z^{-k}dz\\
&=\sum_{k=-\infty}^{+\infty}\frac{f(k)}{1-k}z^{1-k}\\
&=-z\sum_{k=-\infty}^{+\infty}\frac{f(k)}{k-1}z^{-k}\\
\end{align}
Defining $p=k-1$, then $k=p+1$ and the sum can be rewritten as:
\begin{align}
&=-z\sum_{p=-\infty}^{+\infty}\frac{f(p+1)}{p}z^{-p-1}\\
&=-\sum_{p=-\infty}^{+\infty}\frac{f(p+1)}{p}z^{-p}\\
\end{align}
Applying the convolution theorem:
\begin{align}
-\sum_{p=-\infty}^{+\infty}\frac{f(p+1)}{p}z^{-p}&=-\left[\sum_{p=-\infty}^{+\infty}f(p+1)z^{-p} * \sum_{p=-\infty}^{+\infty}\frac{1}{p}z^{-p}\right]\\
\end{align}
Now, applying the time-shift property of the Z-transform:
\begin{align}
-\left[\sum_{p=-\infty}^{+\infty}f(p+1)z^{-p} * \sum_{p=-\infty}^{+\infty}\frac{1}{p}z^{-p}\right]&=-\left[z\sum_{p=-\infty}^{+\infty}f(p)z^{-p} * \sum_{p=-\infty}^{+\infty}\frac{1}{p}z^{-p}\right]\\
&=-z\left[\sum_{p=-\infty}^{+\infty}f(p)z^{-p} * \sum_{p=-\infty}^{+\infty}\frac{1}{p}z^{-p}\right]\\
&=-z\left[\sum_{p=-\infty}^{+\infty}\frac{f(p)}{p}z^{-p}\right]\\
\end{align}
Now, reverting the substitution of $p=k-1$ and $k=p+1$, we get:
\begin{align}
&=-z\left[\sum_{k=-\infty}^{+\infty}\frac{f(k-1)}{k-1}z^{-k+1}\right]\\
&=-z\left[z\sum_{k=-\infty}^{+\infty}\frac{f(k-1)}{k-1}z^{-k}\right]\\
&=-z\left[\sum_{k=-\infty}^{+\infty}\frac{f(k)}{k}z^{-k}\right]\\
&=-z\mathcal{Z}\left\{\frac{f(k)}{k}\right\}
\end{align}
Which means that:
$$\int\mathcal{Z}\{f(k)\}dz=-z\mathcal{Z}\left\{\frac{f(k)}{k}\right\}$$
And, finally:
$$\mathcal{Z}\left\{\frac{f(k)}{k}\right\}=-\frac{1}{z}\int\mathcal{Z}\{f(k)\}dz$$
Which actually differs from the property you posted on your question... If the integral is in $z$, then it makes a difference to have $z^{-1}$ inside or outside the integral.
Hope I was able to help.
