# 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

• Note: in formatting parentheses if you use "\left( " and "\right)" it will scale them up to suit the expression they contain. – lulu Feb 14 at 3:06

If in doubt, write the terms out explicitly ... lets leave the $$1/6$$ out $$\begin{eqnarray*} \sum_{k=8}^{\infty}\left(\frac{5}{6}\right)^{k-1} &=& \left(\frac{5}{6}\right)^{7} + \left(\frac{5}{6}\right)^{8} + \left(\frac{5}{6}\right)^{9} + \cdots \\ &=& \left(\frac{5}{6}\right)^{7} \left(1 + \frac{5}{6} + \left(\frac{5}{6}\right)^{2} + \cdots \right) \\ &=& \left(\frac{5}{6}\right)^{7} \sum_{j=0}^{\infty} \left(\frac{5}{6}\right)^{j}. \\ \end{eqnarray*}$$
\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}