Answer
$\sum\limits_{k=2}^{5}{\frac{k-1}{k+1}}$ is $\frac{21}{10}.$
Work Step by Step
$\sum\limits_{k=2}^{5}{\frac{k-1}{k+1}}$
For the sum of the notation,
$\begin{align}
& \sum\limits_{k=2}^{5}{\frac{k-1}{k+1}}=\frac{2-1}{2+1}+\frac{3-1}{3+1}+\frac{4-1}{4+1}+\frac{5-1}{5+1} \\
& =\frac{1}{3}+\frac{2}{4}+\frac{3}{5}+\frac{4}{6} \\
& =\frac{1}{3}+\frac{1}{2}+\frac{3}{5}+\frac{2}{3}
\end{align}$
And,
$\begin{align}
& \frac{1}{3}+\frac{1}{2}+\frac{3}{5}+\frac{2}{3}=\frac{10+15+18+20}{30} \\
& =\frac{63}{30} \\
& =\frac{21}{10}
\end{align}$
Thus, the sum of the sigma notation $\sum\limits_{k=2}^{5}{\frac{k-1}{k+1}}$ is $\frac{21}{10}.$
