# Solve $\sum nx^n$

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?

• 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 $\frac{d}{dx}\frac{1}{1-x}=\frac{1}{(1-x)^2}$?
– anon
Mar 17 '13 at 22:19
• The methods are the same, but the book one uses derivative notation correctly. Mar 17 '13 at 22:23
• 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 $x^n$).
– anon
Mar 17 '13 at 22:23
• Ok this is definitely my problem. I don't understand the derivative notation. Mar 17 '13 at 22:26
• Possible duplicate of How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$ Dec 13 '16 at 17:39

\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}
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.
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: $$\sum_{n \ge 0} p(n) a_n x^n = p(x D) A(x)$$