Answer
$\sum\limits_{k=3}^{5}{\frac{{{\left( -1 \right)}^{k}}}{k\left( k+1 \right)}}$ is $-\frac{1}{15}\text{.}$
Work Step by Step
$\sum\limits_{k=3}^{5}{\frac{{{\left( -1 \right)}^{k}}}{k\left( k+1 \right)}}$
For the sum of the notation,
$\begin{align}
& \sum\limits_{k=3}^{5}{\frac{{{\left( -1 \right)}^{k}}}{k\left( k+1 \right)}}=\frac{{{\left( -1 \right)}^{3}}}{3\left( 3+1 \right)}+\frac{{{\left( -1 \right)}^{4}}}{4\left( 4+1 \right)}+\frac{{{\left( -1 \right)}^{5}}}{5\left( 5+1 \right)} \\
& =-\frac{1}{12}+\frac{1}{20}-\frac{1}{30}
\end{align}$
And,
$\begin{align}
& \frac{1}{20}-\frac{1}{12}-\frac{1}{30}=\frac{3-5-2}{60} \\
& =\frac{-4}{60} \\
& =-\frac{1}{15}
\end{align}$
Thus, the sum of the sigma notation $\sum\limits_{k=3}^{5}{\frac{{{\left( -1 \right)}^{k}}}{k\left( k+1 \right)}}$ is $-\frac{1}{15}\text{.}$
