Sum of $\sum_{n=1}^\infty n(n+1)x^n$?

Question

$$\sum_{n=1}^\infty n(n+1)x^n$$

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.

https://math.stackexchange.com/questions/593996/how-to-prove-sum-n-0-infty-fracn22n-6/594019#594019 — lab bhattacharjee, Mar 31 '20 at 18:20
https://math.stackexchange.com/questions/593996/how-to-prove-sum-n-0-infty-fracn22n-6/594019#594019 — lab bhattacharjee, Mar 31 '20 at 18:21

score 5 · Answer 1 · answered Mar 31 '20 at 18:21

5

Hint: start from the series for $1/(1-x)$, differentiate, multiply by an appropriate power of $x$, and differentiate again.

answered Mar 31 '20 at 18:21

Robert Israel

448,999

score 3 · Answer 2 · answered Mar 31 '20 at 18:23

This $$ f(x) = \sum_{n=0}^\infty x^n = \frac{1}{1-x} $$ is known. Use it to compute $$ x f(x) = \sum_{n=0}^\infty x^{n+1} \\ (x f(x))' = \sum_{n=0}^\infty (n+1)x^{n} \\ (x f(x))'' = \sum_{n=0}^\infty n(n+1)x^{n-1} \\ x(x f(x))'' = \sum_{n=0}^\infty n(n+1)x^{n} $$

score 0 · Answer 3 · answered Mar 31 '20 at 20:52

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 $|x|<1.$

Sum of $\sum_{n=1}^\infty n(n+1)x^n$?

3 Answers3