Solve $\sum nx^n$

Question

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. — André Nicolas, 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. — dbyrne, Mar 17 '13 at 22:26
Possible duplicate of How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$ — Alex M., Dec 13 '16 at 17:39

score 15 · Accepted Answer · answered Mar 17 '13 at 22:30

Convergence of the series below is assumed throughout.

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

score 7 · Answer 2 · edited Oct 09 '19 at 18:19

7

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.

edited Oct 09 '19 at 18:19

J. W. Tanner

60,406

answered Mar 17 '13 at 22:36

score 4 · Answer 3 · answered Mar 17 '13 at 23:03

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

score 1 · Answer 4 · answered Oct 07 '22 at 02:36

Here's a solution that use neither differentiation nor integraion. Since this question is marked as the duplicate target of How do I compute $\sum_{k=1}^{\infty} k \cdot p^k$, which no longer accepts new solution, I'm posting mine for fun.

Use summation by parts $$S_N = \sum_{n=0}^N a_nb_n = a_N B_N - \sum_{n=0}^{N-1} B_n(a_{n+1}-a_n),$$ where $B_n = \sum_{k=0}^n b_k$ with $a_n = n$ and $b_n = x^n$ for all $n \in \Bbb{N}\cup\{0\}$ and $\lvert x\rvert<1$.

Use geometric sum formula to calculate $S_N$.

\begin{align} B_n &= \sum_{k=0}^n x^k = \frac{1-x^{n+1}}{1-x} \\ a_{n+1}-a_n &= 1 \\ S_N &= N \cdot \frac{1-x^{N+1}}{1-x} - \sum_{n=0}^{N-1} \frac{1-x^{n+1}}{1-x} \\ &= \frac{1}{1-x} \left[(N- Nx^{N+1}) - \sum_{n=0}^{N-1} (1-x^{n+1}) \right] \\ &= \frac{1}{1-x} \left(- Nx^{N+1} + x\sum_{n=0}^{N-1} x^n\right) \\ &= \frac{1}{1-x} \left(- Nx^{N+1} + x \cdot \frac{1-x^N}{1-x} \right) \\ &= \frac{x}{(1-x)^2} - \frac{N(1+x)x^{N+1} + x^{N+1}}{(1-x)^2} \label{eqn} \tag{$\star$} \end{align}

Take $N\to+\infty$ to kill the second term in \eqref{eqn}. Hence $$\boxed{S = \lim_{N\to+\infty} S_N = \frac{x}{(1-x)^2}.}$$

Solve $\sum nx^n$

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