Answer
$1+\dfrac{1}{2}+\dfrac{1}{3}+\dots+\dfrac{1}{99}$
Work Step by Step
We rewrite the given sum by expanding it: in each term $\dfrac{1}{j-1}$ substitute the values of $j$, where $j=2,3,\dots,100$:
$$\begin{align*}
\sum_{j=2}^{10=}\dfrac{1}{j-1}&=\dfrac{1}{2-1}+\dfrac{1}{3-1}+\dfrac{1}{4-1}+\dots+\dfrac{1}{100-1}\\
&=1+\dfrac{1}{2}+\dfrac{1}{3}+\dots+\dfrac{1}{99}.
\end{align*}$$
We used dots in the sum because we cannot list all the $99$ terms of the sum.