Answer
$\sum\limits_{k=2}^{8}{\frac{k}{k-1}}\text{ }$is$\frac{1343}{140}.$
Work Step by Step
$\sum\limits_{k=2}^{8}{\frac{k}{k-1}}$
For the sum of the notation,
$\begin{align}
& \sum\limits_{k=2}^{8}{\frac{k}{k-1}}=\frac{2}{2-1}+\frac{3}{3-1}+\frac{4}{4-1}+\frac{5}{5-1}+\frac{6}{6-1}+\frac{7}{7-1}+\frac{8}{8-1} \\
& =2+\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{8}{7} \\
& =\frac{1343}{140}
\end{align}$
Thus, the sum of the sigma notation $\sum\limits_{k=2}^{8}{\frac{k}{k-1}}$ is $\frac{1343}{140}.$
