Q:

Changing index in summation

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

A:

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*}

A:

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

