How can I find to what this sum converges to?
$$\sum _{n=1}^{\infty} n(n+1)x^n$$
I proved that it converges when
$$|x| < 1$$
but no idea how to find what it sums to.
How can I find to what this sum converges to?
$$\sum _{n=1}^{\infty} n(n+1)x^n$$
I proved that it converges when
$$|x| < 1$$
but no idea how to find what it sums to.
Let $$f(x)=x^1+x^2+x^3+\cdots$$ Then, for $|x|<1$ $$ f'(x)=1+2x+3x^2+4x^3+\cdots $$ $$\eqalign{ f''(x)&=2\cdot1\cdot x^0+3\cdot 2\cdot x+4\cdot3\cdot x^2+\cdots\cr &=\sum_{n=1}^\infty (n+1)n x^{n-1}. } $$ So $$xf''(x)=\sum_{n=1}^\infty (n+1)n x^n.$$
But, for $|x|<1$, $$ f(x)={x\over 1-x} $$ $$f'(x)={d\over dx}{x\over 1-x}={1\over (1-x)^2}$$ $$f''(x)={2\over( 1-x)^3}$$
So $$ {2x\over( 1-x)^3} = \sum_{n=1}^\infty (n+1)n x^n,$$ for $|x|<1$.