## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$10-5+\frac{5}{2}-\ldots$ The limit of the infinite geometric series is given by ${{s}_{\infty }}=\frac{{{a}_{1}}}{1-r}$ When $\left| r \right|<1$ then the limit exists and if $\left| r \right|\ge 1$, no limit exists. To check the limit find the common ratio $r=\frac{{{a}_{n+1}}}{{{a}_{n}}}$ Put $n=1$ in $r=\frac{{{a}_{n+1}}}{{{a}_{n}}}$, \begin{align} & r=\frac{{{a}_{1+1}}}{{{a}_{1}}} \\ & =\frac{{{a}_{2}}}{{{a}_{1}}} \end{align} Put ${{a}_{1}}=10$ and ${{a}_{2}}=-5$ in $r=\frac{{{a}_{2}}}{{{a}_{1}}}$ \begin{align} & r=\frac{{{a}_{2}}}{{{a}_{1}}} \\ & =\frac{-5}{10} \\ & =-\frac{1}{2} \end{align} And, \begin{align} & \left| r \right|=\left| -\frac{1}{2} \right| \\ & =\frac{1}{2} \end{align} Thus, $\left| r \right|=\frac{1}{2}<1$, then the limit exists. Therefore, the given statement is true.