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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?

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Hint:

$$\frac d{dx} \sum_{k=0}^{+\infty} x^k = \sum_{k=1}^{+\infty} kx^{k-1}$$ and $$\frac d{dx} \sum_{k=1}^{+\infty} kx^{k-1} = \sum_{k=2}^{+\infty} (k-1)kx^{k-2}$$

First rewrite your given series as $$\sum_{k=2}^{+\infty} (k-1)k\left(\frac13\right)^{k+1} = \frac1{27} \sum_{k=2}^{+\infty}(k-1)k\left(\frac13\right)^{k-2}$$ and then use the hint above.

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    $\begingroup$ Thanks a lot tilper. If I'm right the answer is 1/4 :) $\endgroup$ – Robinbux May 3 '17 at 19:25
  • $\begingroup$ No problem, glad to help! It appears your answer is correct! $\endgroup$ – tilper May 3 '17 at 19:32
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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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