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I am computing the unilateral z transform of $\nu(n)$. Please help me with the following questions

How can I find the expression for

$\sum_{n=0}^\infty nx^{-n}=x^{-1} + 2x^{-2} + 3x^{-3}+\ldots$

Please show the steps and explain

Please tell the values of x for which the result will hold (Summation will converge)

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    Two months being a member of this site = learn how to properly type mathematics in this site: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – 2017-01-28
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    This question might help you: [How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$](http://math.stackexchange.com/q/30732). And maybe also some of the posts [linked there](http://math.stackexchange.com/questions/linked/30732). Found by looking at [frequent questions in sequences-and-series tag](http://math.stackexchange.com/questions/tagged/sequences-and-series?sort=frequent). – 2017-02-13

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Notice the following:

$$\frac1{1-z}=\sum_{n=0}^\infty z^n$$

take the derivative of both sides:

$$\frac d{dz}\frac1{1-z}=\sum_{n=0}^\infty nz^{n-1}$$

Multiply both sides by $z$.

$$z\cdot\frac d{dz}\frac1{1-z}=\sum_{n=0}^\infty nz^n$$

Now calculate the derivative and let $z=x^{-1}$ to get your series.

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    Although correct, if you don't know some things about power series it is not clear why you can differentiate an infinite sum term by term. – 2017-01-29
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    Thank you sir.... Can you suggest me some sites from where I can practice these kind of series? – 2017-01-29
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    What are the values of z for which the summation would converge? – 2017-01-29
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    @user3237657 It is subtly implied when I use the geometric series. For what values of $z$ does this geometric series converge? – 2017-01-29
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    @user3237657 As per practice, you can try looking through the [sequences-and-series](http://math.stackexchange.com/questions/tagged/sequences-and-series) tag, but as per outside sources I don't really know. – 2017-01-29
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    @SimplyBeautifulArt By intuition I understand that z should be less than 1. Is there any way to deduce this by matematics – 2017-01-29
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    @user3237657 Ratio test... – 2017-01-29
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Start from $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$. Then $\frac{1}{1-x} + \frac{x}{1-x} = 1 + 2x+2x^2+2x^3+...$, $\frac{1}{1-x} + \frac{x}{1-x} + \frac{x^2}{1-x} = 1 + 2x + 3x^2 + 3x^3+3x^4+...$, and in general, $\frac{1}{1-x} \sum_{n=0}^{\infty} x^n = \frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} nx^{n-1}$. Multiply this by $x$ and substitute $x^{-1}$ for $x$ into both sides to get the result you're looking for.