Problem 7. Which number is greater: $\frac{1}{99}$ or

$$
\frac{1}{9903}+\frac{1}{9903+200}+\frac{1}{9903+200+202}+\ldots+\frac{1}{9903+200+202+\ldots+2018} ?
$$
Answer: The first.

Solution. Let $A$ be the second of the given numbers. If we decrease all denominators of the number $A$ by 3, the resulting number $B$ will be greater than $A$:

$$
A<B=\frac{1}{9900}+\frac{1}{9900+200}+\frac{1}{9900+200+202}+\ldots+\frac{1}{9900+200+202+\ldots+2018}
$$

Consider the denominators of the fractions in number $B$. Notice that $9900=99 \cdot 100$. The next denominator is $99 \cdot 100+200=100 \cdot 101$. The next one is $100 \cdot 101+202=101 \cdot 102$, and so on. The last denominator will be $1009 \cdot 1010$. Therefore,

$$
\begin{gathered}
B=\frac{1}{99 \cdot 100}+\frac{1}{100 \cdot 101}+\frac{1}{101 \cdot 102}+\ldots+\frac{1}{1009 \cdot 1010}= \\
=\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+\ldots+\frac{1}{1009}-\frac{1}{1010}=\frac{1}{99}-\frac{1}{1010}<\frac{1}{99}
\end{gathered}
$$

Thus, $A<B<\frac{1}{99}$.

Answer to variant 2: The first.