106. Calculate the sum:

$$
S=\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{99 \cdot 100}
$$
$\triangle$ Let's represent the first term of the sum as $1-\frac{1}{2}$, the second term as $\frac{1}{2}-\frac{1}{3}$, the third term as $\frac{1}{3}-\frac{1}{4}$, and generally, the $k$-th term as

$$
\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}
$$

We get:

$$
S=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{99}-\frac{1}{100}\right)=1-\frac{1}{100}=0.99
$$

Answer: $0.99$.