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This question seems simple but it's been eating me for 20 mins. on a book I seen something like this : $$ \sum_{k=8}^{\infty}\left(\frac{5}{6}\right)^{k-1}\frac{1}{6} = \frac{1}{6}\left(\frac{5}{6}\right)^{7}\sum_{j=0}^{\infty}\left(\frac{5}{6}\right)^{j} $$ Could anyone tell me why the above step is valid, I understand the $\frac{1}{6}$ part but not the $\left(\frac{5}{6}\right)^{7}$ part. It is from a probability book so conceptually I know this gives the right answer. please help, thank you very much

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    $\begingroup$ Note: in formatting parentheses if you use "\left( " and "\right)" it will scale them up to suit the expression they contain. $\endgroup$ – lulu Feb 14 at 3:06
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If in doubt, write the terms out explicitly ... lets leave the $1/6$ out \begin{eqnarray*} \sum_{k=8}^{\infty}\left(\frac{5}{6}\right)^{k-1} &=& \left(\frac{5}{6}\right)^{7} + \left(\frac{5}{6}\right)^{8} + \left(\frac{5}{6}\right)^{9} + \cdots \\ &=& \left(\frac{5}{6}\right)^{7} \left(1 + \frac{5}{6} + \left(\frac{5}{6}\right)^{2} + \cdots \right) \\ &=& \left(\frac{5}{6}\right)^{7} \sum_{j=0}^{\infty} \left(\frac{5}{6}\right)^{j}. \\ \end{eqnarray*}

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  • $\begingroup$ OMG, Thank you very much !!!! I feel so dumb right now, and I guess adding 8 terms to infinity doesn't change a thing :) $\endgroup$ – A.Y Feb 14 at 3:07
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Breaking it down to simpler steps.

$$\begin{align} \sum_{k=8}^{\infty}\left(\frac{5}{6}\right)^{k-1}\frac{1}{6} ~&=~ \frac{1}{6}\left(\frac{5}{6}\right)^{7}\sum_{k=8}^{\infty}\left(\frac{5}{6}\right)^{k-8}&&\text{distributing out constant factors} \\[1ex]&=~ \frac{1}{6}\left(\frac{5}{6}\right)^{7}\sum_{k-8=0}^{\infty}\left(\frac{5}{6}\right)^{k-8}&&\text{preparing for substitution} \\[1ex]&=~ \frac{1}{6}\left(\frac{5}{6}\right)^{7}\sum_{j=0}^{\infty}\left(\frac{5}{6}\right)^{j}&&\text{substituting }j\text{ for }k-8 \end{align}$$

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