Its been a long tie since I dealt with serieses like this $\sum_{x=1}^\infty xq^{2x}$, I tried to take to a form where I can calculate the derivative of other sum, but didn't work. I'd be happy if someone shows the way.
Note: $|q|<1$
how to calculate $\sum_{x=1}^\infty xq^{2x}$?
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sequences-and-series
2 Answers
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Hint:
$$y\cdot\frac d{dy}\sum_{n=1}^\infty y^n=\sum_{n=1}^\infty ny^n$$
Let $y=q^2$ and apply the geometric series.
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0This is the way I would have proceeded, but since you beat me to it, I posted a solution that uses another way forward. ;-)) – 2017-02-24
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0@Dr.MV I suppose if anyone's interested, a third way could be the binomial theorem ;-) – 2017-02-24
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I thought it might be instructive to present an approach that relies on elementary tools and not on differentiation of a series.
Using $x=\sum_{y=1}^x (1)$, we can write
$$\begin{align} \sum_{x=1}^\infty xq^{2x}&=\sum_{x=1}^\infty \sum_{y=1}^x q^{2x}\\\\ &=\sum_{y=1}^\infty \sum_{x=y}^\infty q^{2x}\\\\ &=\sum_{y=1}^\infty \frac{q^{2y}}{1-q^2}\\\\ &=\frac{q^2}{(1-q^2)^2} \end{align}$$
Tools Used: Interchanging order of summation (Fubini) and Summing Geometric Series.