Infinite Series 1

So we can rewrite the series this way:

$\sum_{n=0}^{\infty} n(n-2)\frac{1}{3^{n-3}}$ = $\sum_{n=0}^{\infty} n(n-2)x^{n-3}$ , where $x=\frac{1}{3}$ , and so noticing the derivative with respect to x:

$\sum_{n=0}^{\infty} n(n-2)x^{n-3}$ = $(\sum_{n=0}^{\infty} nx^{n-2})'$ = $(\frac{1}{x}\sum_{n=0}^{\infty} nx^{n-1})'$ repeting the same process, i.e:

$(\frac{1}{x}\sum_{n=0}^{\infty} nx^{n-1})'$ = $(\frac{1}{x}(\sum_{n=0}^{\infty} x^{n})')'$ and solving the inner part you obtain:

$(\frac{1}{x}(\sum_{n=0}^{\infty} x^{n})')'$ = $(\frac{1}{x}(\frac{1}{1-x})')'$ = $(\frac{1}{x}\frac{1}{(1-x)^2})'$ = $\frac{1-3x}{x^2(x-1)^3}$ , finally as $x=\frac{1}{3}$ you get that:

$\sum_{n=0}^{\infty} n(n-2)\frac{1}{3^{n-3}}$ = 0