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