Answer
$\sum\limits_{k=1}^{\infty}\frac{1}{k(k+1)^2}$
Work Step by Step
We have to rewrite the sum using sigma notation
$\frac{1}{1\cdot2^2}+\frac{1}{2\cdot3^2}+\frac{1}{3\cdot4^2}+\frac{1}{4\cdot5^2}+\cdots$
Each term has the form of $\frac{1}{k(k+1)^2}$, where $k=1,2,3,4,\cdots.$
The number of terms is infinite and the first term is $\frac{1}{1(1+1)^2}$, hence the index of the sum goes from $1$ to $\infty$.
So in sigma notation:
$\frac{1}{1\cdot2^2}+\frac{1}{2\cdot3^2}+\frac{1}{3\cdot4^2}+\frac{1}{4\cdot5^2}+\cdots=\sum\limits_{k=1}^{\infty}\frac{1}{k(k+1)^2}$
