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

We have

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

With the first equality, we rewrote the geometries series starting from $n=0$ in order to apply $\sum _{n=0}^{\mathrm{\infty }}{r}^n=\frac{1}{1-r}$ 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.