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I was recently watching a Numberphile Video when I saw this 'formula' :

$$S = \sum \limits_{n=1} ^ {\infty} n\,x^{n-1} = \dfrac {1} {\big(x-1\big)^2} \,\,\,\forall\,\,\, |x| < 1$$

I was amazed to see this and it immediately struck to my mind that it is really an infinite derivative series:

$$S=\sum \limits_{n=1}^{\infty}\dfrac {d}{dx} x^n = \dfrac {d}{dx}\sum \limits_{n=1}^{\infty}x^n$$

Now applying the sum for infinite Geometric Series when $|r|<1$ ($r = $ common ratio) :

$$S = \dfrac{d}{dx}\dfrac{1}{1-x} = \dfrac{1}{\big(1-x\big)^2}$$

I am really a beginner at Calculus, so the first question I want to ask is :

Is my derivation for the sum $S$ mathematically correct ?

Though my final expression comes out good, there still might be chance of some mistake, so please help me.

Also,

Is there any other way of deriving this result besides Calculus? Is this result even linked to Calculus in any way?

I can clearly see that the common ratio $r$ for $$ is $r=\dfrac{n+1}{n}x$, but unfortunately $n$ is variable and not a fixed constant and also $|r| < 1$ does not necessarily hold for me to be able to compute the infinte sum.

Can anyone help ?

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    The main "problem" here is to argue that you can interchange the derivate with the inifinte summation. – 2017-02-13
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    @NiklasHebestreit Sorry but I do not understand what you want to convey.... – 2017-02-13
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    you can also expand the rhs into a tayloer series... – 2017-02-13
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    Of course this is linked to calculus, it is a common [Maclaurin series expansion](https://en.wikipedia.org/wiki/Taylor_series#List_of_Maclaurin_series_of_some_common_functions). – 2017-02-13
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    For $|x|<1$ we have the [negative binomial expansion](https://brilliant.org/wiki/negative-binomial-theorem/)$$(1-x)^{-2}=1+2x+\frac{2(2+1)x^2}{2!}+\frac{2(2+1)(2+2)}{3!}x^3+...$$ $$=\sum_{k=0}^{\infty}\binom{-2}{k}(-x)^k$$ – 2017-02-13

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By elementary means:

$$S:=\sum_{n=1}^\infty nx^{n-1}=\sum_{n=0}^\infty(n+1)x^n=\sum_{n=0}^\infty nx^n+\sum_{n=0}^\infty x^n=xS+\frac1{1-x},$$ so that

$$S=\frac1{(1-x)^2}.$$

You can also discuss convergence by evaluating the partial sums and showing that the remainder decreases exponentially.

$$S_m:=\sum_{n=1}^m nx^{n-1}=\sum_{n=0}^{m-1}(n+1)x^n=\sum_{n=0}^{m-1} nx^n+\sum_{n=0}^{m-1}x^n=xS_m-mx^m+\frac{1-x^m}{1-x},$$

$$S_m=-\frac{mx^m}{(1-x)}+\frac{1-x^m}{(1-x)^2}=S-\frac{mx^m}{(1-x)}-\frac{x^m}{(1-x)^2}.$$