Sum of $\sum_{n=1}^\infty n(n+1)x^n$? $$\sum_{n=1}^\infty n(n+1)x^n$$
I feel like this is a Taylor series (or the derivative/integral of one), but I'm struggling to come up with the right one. I can see that the interval of convergence is $-1 \cup 1$, but the sum itself escapes me.
 A: Hint: start from the series for $1/(1-x)$, differentiate, multiply by an appropriate power of $x$, and differentiate again.
A: This
$$
f(x) = \sum_{n=0}^\infty x^n = \frac{1}{1-x}
$$
is known.  Use it to compute
$$
x f(x) = \sum_{n=0}^\infty x^{n+1}
\\
(x f(x))' = \sum_{n=0}^\infty (n+1)x^{n}
\\
(x f(x))'' = \sum_{n=0}^\infty n(n+1)x^{n-1}
\\
x(x f(x))'' = \sum_{n=0}^\infty n(n+1)x^{n}
$$
A: You can do summation by parts twice to find the sum from $n=1$ to $n=m$ and then take the limit as $m \to \infty.$ I'll use the notaion $\Delta g(n)=g(n+1)-g(n)$ and the formula $$\sum_{n=1}^mf(n)\Delta g(n)=f(n)g(n)|_{n=1}^{m+1}-\sum_{n=1}^m(\Delta f(n)) g(n+1).$$ I'll let you do the algebra, noting that $x^n=\Delta (\frac{x^n}{x-1})$, that $\Delta n(n+1)=2(n+1)$, that $\Delta (n+1)=1$ and that $x$ and $\frac {1}{x-1}$ are constants and that constants can be moved outside the $\Delta$ operator and the summation symbol $\Sigma.$ When taking the limit as $m \to \infty$, use $\lim_{m\to \infty}x^m=0$ for $|x|<1.$
