Answer
$\sum_{k=1}^{999} \dfrac{1}{k(k+1)}$
Work Step by Step
Writing the given sum using sigma notation means to determine the general term of the sum and the limits for the summation index:
$$\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+\dots+\dfrac{1}{999\cdot 1000}=\sum_{k=m}^n a_k.$$
We look for the pattern in the terms of the sum:
$$\begin{align*}
\dfrac{1}{1\cdot 2}&=\dfrac{1}{1\cdot (1+1)}\\
\dfrac{1}{2\cdot 3}&=\dfrac{1}{2\cdot (2+1)}\\
\dfrac{1}{3\cdot 4}&=\dfrac{1}{3\cdot (3+1)}\\
&\dots\\
\dfrac{1}{999\cdot 1000}&=\dfrac{1}{999\cdot (999+1)}.
\end{align*}$$
So the term on the $k$th position is written as
$$a_k=\dfrac{1}{k\cdot (k+1)} .$$
The summation index $k$ goes from $1$ to $999$.
We conclude that the sum can be written using sigma notation like this:
$$\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+\dots+\dfrac{1}{999\cdot 1000}=\sum_{k=1}^{999} \dfrac{1}{k(k+1)}.$$
