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I am trying to find a closed form expression for $\sum_{i=1}^{\infty}i(2y^i-y^{2i})^b$ where $0<y<1$ and b is a positive integer. I was thinking binomial expansion but didn't get anywhere with it.

Any help appreciated.

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  • $\begingroup$ Note that the series diverges if $b=0$. $\endgroup$ – Servaes Apr 25 at 20:34
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Using the binomial theorem: $$ \begin{align} & \quad \quad \quad =\sum\nolimits_{i=1}^{\infty }{\left[ i\sum\nolimits_{k=0}^{b}{\left( \begin{matrix} b \\ k \\ \end{matrix} \right){{2}^{k}}{{y}^{ik}}{{\left( -{{y}^{2i}} \right)}^{b-k}}} \right]} \\ & \quad \quad \quad =\sum\nolimits_{i=1}^{\infty }{\left[ i\sum\nolimits_{k=0}^{b}{{{\left( -1 \right)}^{b-k}}{{2}^{k}}\left( \begin{matrix} b \\ k \\ \end{matrix} \right){{y}^{2ib-ik}}} \right]} \\ & \quad \quad \quad =\sum\nolimits_{i=1}^{\infty }{\left[ \sum\nolimits_{k=0}^{b}{c\left( a,k \right)i{{y}^{2ib-ik}}} \right]}\quad \quad where\quad c\left( a,k \right)={{\left( -1 \right)}^{b-k}}{{2}^{k}}\left( \begin{matrix} b \\ k \\ \end{matrix} \right) \\ & \quad \quad \quad =\sum\nolimits_{k=0}^{b}{\left[ c\left( a,k \right)\sum\nolimits_{i=1}^{\infty }{i{{y}^{2ib-ik}}} \right]} \\ & \quad \quad \quad =\sum\nolimits_{k=0}^{b}{\left[ c\left( a,k \right)\frac{{{y}^{2b+k}}}{{{\left( {{y}^{k}}-{{y}^{2b}} \right)}^{2}}} \right]} \\ \end{align} $$ in the last line i used the identity(for a positive integer$ \alpha$): $$\sum\nolimits_{i=1}^{\infty }{i{{y}^{i\alpha }}}=\frac{{{y}^{\alpha }}}{{{\left( {{y}^{\alpha }}-1 \right)}^{2}}}$$

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    $\begingroup$ I was aware that binomial expansion would yield another sum, although with finite indices. However, my goal is to find a closed form expression for this sum if possible. $\endgroup$ – Zaeem Hussain Apr 26 at 10:12

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