<s>[INST] How to get the limit of this series I need to solve this series:
$$\sum _{ k=2 }^{ \infty  } (k-1)k \left( \frac{ 1 }{ 3 }   \right) ^{ k+1 }$$
I converted it into $$\sum _{ k=0 }^{ \infty  } \frac { { k }^{ 2 }-k }{ 3 } \left(\frac { 1 }{ 3 }   \right)^{ k } -\frac { 4 }{ 3 } $$ with the idea, that $$\sum _{ k=0 }^{ \infty  }{ { q }^{ k } } =\frac { 1 }{ 1-q } $$but I don't know how to get rid of $\frac { { k }^{ 2 }-k }{ 3 }$.
Can someone please tell me how to go on?
 [/INST] By stars and bars the coefficient of $x^n$ in $\frac{1}{(1-x)^3}$, that is the number of ways of writing $n$ as the sum of three natural numbers (or the number of ways for writing $n+3$ as the sum of three positive natural numbers), equals $\binom{n+2}{2}$. It follows that
$$ \frac{1}{(1-x)^3} = \sum_{n\geq 0}\binom{n+2}{2}x^n,\qquad \frac{x^2}{(1-x)^3}=\sum_{m\geq 2}\binom{m}{2}x^m $$
and by evaluating $\frac{2x^2}{3(1-x)^3}$ at $x=\frac{1}{3}$ it follows that
$$ \sum_{k\geq 2}k(k-1)\frac{1}{3^{k+1}}=\color{red}{\frac{1}{4}}.$$
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