Science:Math Exam Resources/Courses/MATH105/April 2015/Question 01 (l)/Solution 1

Following the hint, we rewrite our series as follows:

$\sum _{k=7}^{\infty }\left({\frac {1}{8}}\right)^{k}=\sum _{k=0}^{\infty }\left({\frac {1}{8}}\right)^{k}-\sum _{k=0}^{6}\left({\frac {1}{8}}\right)^{k}.$

The first series on the right hand side is a familiar infinite geometric series of the form $\sum _{k=0}^{\infty }c^{k}$ that converges to ${\frac {1}{1-c}}$ for $|c|<1.$

Therefore,

$\sum _{k=0}^{\infty }\left({\frac {1}{8}}\right)^{k}={\frac {1}{1-1/8}}={\frac {1}{7/8}}={\color {red}{\frac {8}{7}}}.$

The second series on the right hand side of the first equation is a finite geometric series of the form $\sum _{k=0}^{n}ar^{k}$ that converges to $a\cdot {\frac {1-r^{n+1}}{1-r}}.$

In our case, $r={\frac {1}{8}}$ and $a=1$; therefore,

$\sum _{k=0}^{6}\left({\frac {1}{8}}\right)^{k}={\frac {1-\left({\frac {1}{8}}\right)^{7}}{1-{\frac {1}{8}}}}={\color {red}{\frac {8(1-({\frac {1}{8}})^{7})}{7}}}.$

Finally, combining everything together yields

$\sum _{k=7}^{\infty }\left({\frac {1}{8}}\right)^{k}=\sum _{k=0}^{\infty }\left({\frac {1}{8}}\right)^{k}-\sum _{k=0}^{6}\left({\frac {1}{8}}\right)^{k}={\frac {8}{7}}-{\frac {8(1-\left({\frac {1}{8}}\right)^{7})}{7}}={\frac {1}{7}}\cdot {\frac {8}{8^{7}}}={\color {blue}{\frac {1}{7\cdot 8^{6}}}}.$