Science:Math Exam Resources/Courses/MATH105/April 2014/Question 01 (l)/Solution 1

We have

${\begin{aligned}\sum _{n=1}^{\infty }\left[\left({\frac {1}{3}}\right)^{n}+\left(-{\frac {2}{5}}\right)^{n-1}\right]&=\sum _{n=0}^{\infty }\left[\left({\frac {1}{3}}\right)^{n+1}+\left(-{\frac {2}{5}}\right)^{n}\right]\\&={\frac {1}{3}}\sum _{n=0}^{\infty }\left({\frac {1}{3}}\right)^{n}+\sum _{n=0}^{\infty }\left(-{\frac {2}{5}}\right)^{n}\\&={\frac {1}{3}}{\frac {1}{1-1/3}}+{\frac {1}{1-(-2/5)}}\\&={\frac {1}{2}}+{\frac {5}{7}}\\&={\frac {17}{14}}.\end{aligned}}$

With the first equality, we rewrote the geometries series starting from $n=0$ in order to apply $\sum _{n=0}^{\infty }r^{n}={\frac {1}{1-r}}$ for $|r|<1$. This was accomplished by re-indexing. In the second equality, we split the sum into two and factored out $1/3$ from the first sum.