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I am trying to find a closed form solution for $\sum_{n\ge0} nx^n\text{, where }\lvert x \rvert<1$.

This solution makes sense to me:

$\sum_{n\ge0} x^n=(1-x)^{-1} \\ \frac{d}{d x} \sum_{n\ge0} x^n = \frac{d}{d x} (1-x)^{-1} \\ \sum_{n\ge0} nx^{n-1} = (1-x)^{-2} \\ x \sum_{n\ge0} n x^{n-1} = x(1-x)^{-2} \\ \sum_{n\ge0} nx^n=\frac x{(1-x)^2}$

However, a book I am reading used the following method:

$$\sum_{n\ge0}nx^n=\sum_{n\ge0}x\frac d{dx}x^n= x\frac d{dx}\sum\limits_{n\ge0}x^n=x\frac d{dx}\frac1{1-x}=\frac x{(1-x)^2}$$

This seems closely related to the solution I described above, but I am having difficulty understanding it. Can someone explain the method being used here?

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    $\begingroup$ Which equality specifically are you having trouble understanding? Do you agree that $nx^n=x\frac{d}{dx}x^n$? Do you agree that $x$ can be factored out of the sum? Do you agree that $d/dx$ can be factored out? Do you agree that $\sum_0x^n=\frac{1}{1-x}$? Do you agree that $\frac{d}{dx}\frac{1}{1-x}=\frac{1}{(1-x)^2}$? $\endgroup$
    – anon
    Mar 17 '13 at 22:19
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    $\begingroup$ The methods are the same, but the book one uses derivative notation correctly. $\endgroup$ Mar 17 '13 at 22:23
  • $\begingroup$ In your solution, why are you putting $\frac{d}{dx}$ on the right side of everything? Putting differential operators on the right is understood in the sense of operator algebras, where $x^n\frac{d}{dx}$ and $nx^{n-1}$ are not the same: the second expression $nx^{n-1}$ is the derivative of $x^n$ (that is, $\frac{d}{dx}x^n=nx^{n-1}$), whereas the first expression $x^n\frac{d}{dx}$ is an operator which sends a function $f(x)$ to $x^nf'(x)$ (i.e. take the derivative of $f$, then multiply by $x^n$). $\endgroup$
    – anon
    Mar 17 '13 at 22:23
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    $\begingroup$ Ok this is definitely my problem. I don't understand the derivative notation. $\endgroup$
    – dbyrne
    Mar 17 '13 at 22:26
  • $\begingroup$ Possible duplicate of How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$ $\endgroup$
    – Alex M.
    Dec 13 '16 at 17:39
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Convergence of the series below is assumed throughout.

$$\begin{align} \sum\limits_{n\ge0}nx^n&=\sum\limits_{n\ge0}x(x^n)' &\text{integrate } nx^{n-1}\\ &=x\sum\limits_{n\ge0}(x^n)' &\text{factor } x \,\text{out}\\ &=x\left(\sum\limits_{n\ge0}(x^n)\right)' &\text{differentiate the whole series} \\ &=x\left(\frac1{1-x}\right)' &|x|<1\\ &=\frac x{(1-x)^2} &\text{differentiate }\frac {1}{1-x} \end{align}$$

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Hint: The basic idea that we can switch $\frac{d}{dx}$ and $\sum$ in any compact subset of the disc of convergence for the power series.

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Both your and the book's development are the same. The advantage of the book's method is that it points at the following: If you want terms in $n x^n$, you get them by $x \dfrac{d}{dx} x^n$, to get $n^2 x^n$, you do $x \dfrac{d}{d x} \left(x \dfrac{d}{d x} x^n \right)$, and so on. If you use the notation $D$ for the operator $\dfrac{d}{d x}$, you can then write $n x^n = x D x^n$, $n^2 x^n = (x D)^2 x^n$, and in general $n^k x^n = (x D)^k x^n$, and if now $p(\cdot)$ is any polynomial, by combining several of the previous formulas you get $p(n) x^n = p(x D) x^n$.

Let $A(x) = \sum_{n \ge 0} a_n x^n$. This idea applied term by term to $A(x)$ is: $$ \sum_{n \ge 0} p(n) a_n x^n = p(x D) A(x) $$

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