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How do I find the following sum? $$\sum_{x=3}^\infty 1.536x\left(\frac{5}{8}\right)^x$$ It wouldn't be geometric because of the $x$ in front, right?

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Hint: You are correct that it's not a geometric series. What we can do is note that

$$\sum\limits_{n = 0}^{\infty} r^n = \frac{1}{1 - r}$$

Now differentiating termwise, we see that

$$\sum\limits_{n = 1}^{\infty} nr^{n - 1} = \frac{1}{(1 - r)^2}$$

This can be adapted for your series:

\begin{align} \sum\limits_{x = 3}^{\infty} x \left(\frac 5 8 \right)^x &= \frac 5 8 \sum\limits_{x = 3}^{\infty} x \left(\frac 5 8\right)^{x - 1} \\ &= \frac 5 8 \left(\sum\limits_{x = 1}^{\infty} x \left(\frac 5 8 \right)^{x - 1} - 1 - 2 \cdot \frac 5 8\right) \\ &= \frac 5 8 \left(\frac{1}{(1 - (5/8))^2}-1 - \frac {10} 8\right) \end{align}

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  • $\begingroup$ Ahh, that makes sense now. You subtract the first 2 terms because it starts at 3 right? $\endgroup$ – Jesus Oct 14 '13 at 1:23
  • $\begingroup$ @Jesus Exactly. $\endgroup$ – user61527 Oct 14 '13 at 1:23

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