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# Question from DragonLord

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Question from DragonLord :

$$\sum \limits_{n=1}^{\infty} \dfrac{n^3}{3^n} = \ ?$$

My attempt:

$$\begin{array}{|rcll|} \hline s &=& \sum \limits_{n=1}^{\infty} \dfrac{n^3}{3^n} \\ \dfrac{1}{3}s &=& \sum \limits_{n=1}^{\infty} \dfrac{n^3}{3^{n+1}} \\ \hline s-\dfrac{1}{3}s &=& \sum \limits_{n=1}^{\infty} \dfrac{n^3}{3^n} -\sum \limits_{n=1}^{\infty} \dfrac{n^3}{3^{n+1}} \\ \dfrac{2}{3}s &=& \dfrac{1}{3}+ \sum \limits_{n=2}^{\infty} \dfrac{n^3}{3^n} -\sum \limits_{n=1}^{\infty} \dfrac{n^3}{3^{n+1}} \\ \dfrac{2}{3}s &=& \dfrac{1}{3}+ \sum \limits_{n=1}^{\infty} \dfrac{(n+1)^3}{3^{n+1}} -\sum \limits_{n=1}^{\infty} \dfrac{n^3}{3^{n+1}} \\ \dfrac{2}{3}s &=& \dfrac{1}{3}+ \sum \limits_{n=1}^{\infty} \dfrac{n^3+3n^2+3n+1}{3^{n+1}} -\sum \limits_{n=1}^{\infty} \dfrac{n^3}{3^{n+1}} \\ \dfrac{2}{3}s &=& \dfrac{1}{3}+\sum \limits_{n=1}^{\infty} \dfrac{n^3}{3^{n+1}} + \sum \limits_{n=1}^{\infty} \dfrac{3n^2+3n+1}{3^{n+1}} -\sum \limits_{n=1}^{\infty} \dfrac{n^3}{3^{n+1}} \\ \dfrac{2}{3}s &=& \dfrac{1}{3}+\sum \limits_{n=1}^{\infty} \dfrac{3n^2+3n+1}{3^{n+1}} \\ \dfrac{2}{3}s &=& \dfrac{1}{3}+\dfrac{1}{3}\sum \limits_{n=1}^{\infty} \dfrac{3n^2+3n+1}{3^{n}} \\ \dfrac{2}{3}s &=& \dfrac{1}{3}+\dfrac{1}{3}\sum \limits_{n=1}^{\infty} \dfrac{3n^2}{3^{n}} +\dfrac{1}{3}\sum \limits_{n=1}^{\infty} \dfrac{3n}{3^{n}} +\dfrac{1}{3}\sum \limits_{n=1}^{\infty} \dfrac{1}{3^{n}} \\ \dfrac{2}{3}s &=& \dfrac{1}{3}+\dfrac{3}{3}\sum \limits_{n=1}^{\infty} \dfrac{n^2}{3^{n}} +\dfrac{3}{3}\sum \limits_{n=1}^{\infty} \dfrac{n}{3^{n}} +\dfrac{1}{3}\sum \limits_{n=1}^{\infty} \left( \dfrac{1}{3}\right)^{n} \\ \dfrac{2}{3}s &=& \dfrac{1}{3}+\dfrac{3}{3}\sum \limits_{n=1}^{\infty} \dfrac{n^2}{3^{n}} +\dfrac{3}{3}\sum \limits_{n=1}^{\infty} \dfrac{n}{3^{n}} +\dfrac{1}{3}\left( \dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}} \right) \\ \dfrac{2}{3}s &=& \dfrac{1}{3}+\dfrac{3}{3}\sum \limits_{n=1}^{\infty} \dfrac{n^2}{3^{n}} +\dfrac{3}{3}\sum \limits_{n=1}^{\infty} \dfrac{n}{3^{n}} +\dfrac{1}{6} \\ \dfrac{2}{3}s &=&\dfrac{1}{3}+\dfrac{1}{6} +\dfrac{3}{3}\sum \limits_{n=1}^{\infty} \dfrac{n^2}{3^{n}} +\dfrac{3}{3}\sum \limits_{n=1}^{\infty} \dfrac{n}{3^{n}} \\\\ \mathbf{\dfrac{2}{3}s} &=&\mathbf{ \dfrac{1}{3}+\dfrac{1}{6} + \sum \limits_{n=1}^{\infty} \dfrac{n^2}{3^{n}} + \sum \limits_{n=1}^{\infty} \dfrac{n}{3^{n}} } \qquad (1)\\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \sum \limits_{n=1}^{\infty} \dfrac{n}{3^{n}} &=& \ ? \\ \hline s_2 &=& \sum \limits_{n=1}^{\infty} \dfrac{n}{3^{n}} \\ \dfrac{1}{3}s_2 &=& \sum \limits_{n=1}^{\infty} \dfrac{n}{3^{n+1}} \\ \hline s_2-\dfrac{1}{3}s_2 &=& \sum \limits_{n=1}^{\infty} \dfrac{n}{3^{n}} -\sum \limits_{n=1}^{\infty} \dfrac{n}{3^{n+1}} \\ \dfrac{2}{3}s_2 &=& \dfrac{1}{3}+ \sum \limits_{n=2}^{\infty} \dfrac{n}{3^{n}} -\sum \limits_{n=1}^{\infty} \dfrac{n}{3^{n+1}} \\ \dfrac{2}{3}s_2 &=& \dfrac{1}{3}+ \sum \limits_{n=1}^{\infty} \dfrac{n+1}{3^{n+1}} -\sum \limits_{n=1}^{\infty} \dfrac{n}{3^{n+1}} \\ \dfrac{2}{3}s_2 &=& \dfrac{1}{3}+ \sum \limits_{n=1}^{\infty} \dfrac{n+1-n}{3^{n+1}} \\ \dfrac{2}{3}s_2 &=& \dfrac{1}{3}+ \sum \limits_{n=1}^{\infty} \dfrac{1}{3^{n+1}} \\ \dfrac{2}{3}s_2 &=& \dfrac{1}{3}+ \dfrac{1}{3}\sum \limits_{n=1}^{\infty} \dfrac{1}{3^{n}} \\ \dfrac{2}{3}s_2 &=& \dfrac{1}{3}+ \dfrac{1}{3}\sum \limits_{n=1}^{\infty} \left(\dfrac{1}{3}\right)^{n} \\ \dfrac{2}{3}s_2 &=& \dfrac{1}{3}+\dfrac{1}{3}\left( \dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}} \right) \quad | \quad * 3 \\ 2s_2 &=& 1+ \left( \dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}} \right) \\ 2s_2 &=& 1+ \dfrac{1}{2} \\ 2s_2 &=& \dfrac{3}{2} \\ s_2 &=& \dfrac{3}{4} \\\\ \mathbf{ \sum \limits_{n=1}^{\infty} \dfrac{n}{3^{n}} } &=& \mathbf{ \dfrac{3}{4} } \qquad (2) \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \sum \limits_{n=1}^{\infty} \dfrac{n^2}{3^{n}} &=& \ ? \\ \hline s_3 &=& \sum \limits_{n=1}^{\infty} \dfrac{n^2}{3^{n}} \\ \dfrac{1}{3}s_3 &=& \sum \limits_{n=1}^{\infty} \dfrac{n^2}{3^{n+1}} \\ \hline s_3-\dfrac{1}{3}s_3 &=& \sum \limits_{n=1}^{\infty} \dfrac{n^2}{3^{n}} -\sum \limits_{n=1}^{\infty} \dfrac{n^2}{3^{n+1}} \\ \dfrac{2}{3}s_3 &=& \sum \limits_{n=1}^{\infty} \dfrac{n^2}{3^{n}} -\sum \limits_{n=1}^{\infty} \dfrac{n^2}{3^{n+1}} \\ \dfrac{2}{3}s_3 &=& \dfrac{1}{3}+ \sum \limits_{n=2}^{\infty} \dfrac{n^2}{3^{n}} -\sum \limits_{n=1}^{\infty} \dfrac{n^2}{3^{n+1}} \\ \dfrac{2}{3}s_3 &=& \dfrac{1}{3}+ \sum \limits_{n=1}^{\infty} \dfrac{(n+1)^2}{3^{n+1}} -\sum \limits_{n=1}^{\infty} \dfrac{n^2}{3^{n+1}} \\ \dfrac{2}{3}s_3 &=& \dfrac{1}{3}+ \sum \limits_{n=1}^{\infty} \dfrac{(n+1)^2-n^2}{3^{n+1}} \\ \dfrac{2}{3}s_3 &=& \dfrac{1}{3}+ \sum \limits_{n=1}^{\infty} \dfrac{(n^2+2n+1-n^2}{3^{n+1}} \\ \dfrac{2}{3}s_3 &=& \dfrac{1}{3}+ \sum \limits_{n=1}^{\infty} \dfrac{2n+1}{3^{n+1}} \\ \dfrac{2}{3}s_3 &=& \dfrac{1}{3}+ \sum \limits_{n=1}^{\infty} \dfrac{2n}{3^{n+1}} + \sum \limits_{n=1}^{\infty} \dfrac{1}{3^{n+1}} \\ \dfrac{2}{3}s_3 &=& \dfrac{1}{3}+ \dfrac{2}{3}\sum \limits_{n=1}^{\infty} \dfrac{n}{3^{n}} + \dfrac{1}{3}\sum \limits_{n=1}^{\infty} \dfrac{1}{3^{n}} \\ \dfrac{2}{3}s_3 &=& \dfrac{1}{3}+ \dfrac{2}{3}\sum \limits_{n=1}^{\infty} \dfrac{n}{3^{n}} + \dfrac{1}{3}\sum \limits_{n=1}^{\infty} \left(\dfrac{1}{3}\right)^{n} \quad | \quad \sum \limits_{n=1}^{\infty} \dfrac{n}{3^{n}} = \dfrac{3}{4} \\ \dfrac{2}{3}s_3 &=& \dfrac{1}{3}+ \dfrac{2}{3}*\dfrac{3}{4} + \dfrac{1}{3}\sum \limits_{n=1}^{\infty} \left(\dfrac{1}{3}\right)^{n} \quad | \quad \sum \limits_{n=1}^{\infty} \left(\dfrac{1}{3}\right)^{n} = \dfrac{1}{2} \\ \dfrac{2}{3}s_3 &=& \dfrac{1}{3}+ \dfrac{2}{3}*\dfrac{3}{4} + \dfrac{1}{3}*\dfrac{1}{2} \\ \dfrac{2}{3}s_3 &=& \dfrac{1}{3}+ \dfrac{1}{2} + \dfrac{1}{6} \\ \dfrac{2}{3}s_3 &=& \dfrac{5}{6}+ \dfrac{1}{6} \\ \dfrac{2}{3}s_3 &=& 1 \\ s_3 &=& \dfrac{3}{2} \\\\ \mathbf{ \sum \limits_{n=1}^{\infty} \dfrac{n^2}{3^{n}} } &=& \mathbf{ \dfrac{3}{2} } \qquad (3) \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline (1): & \mathbf{\dfrac{2}{3}s} &=&\mathbf{ \dfrac{1}{3}+\dfrac{1}{6} + \sum \limits_{n=1}^{\infty} \dfrac{n^2}{3^{n}} + \sum \limits_{n=1}^{\infty} \dfrac{n}{3^{n}} } \quad | \quad \mathbf{ \sum \limits_{n=1}^{\infty} \dfrac{n^2}{3^{n}} =\dfrac{3}{2} },\quad \mathbf{ \sum \limits_{n=1}^{\infty} \dfrac{n}{3^{n}}=\dfrac{3}{4} } \\\\ & \dfrac{2}{3}s &=& \dfrac{1}{3}+\dfrac{1}{6} + \dfrac{3}{2} + \dfrac{3}{4} \\\\ & \dfrac{2}{3}s &=& \dfrac{11}{4} \\\\ & s &=& \dfrac{3}{2}*\dfrac{11}{4} \\\\ & s &=& \dfrac{33}{8} \\\\ & \mathbf{\sum \limits_{n=1}^{\infty} \dfrac{n^3}{3^n}} &=& \mathbf{\dfrac{33}{8}} \\ \hline \end{array}$$

Mar 25, 2020