Problem 2. a) Determine the natural numbers $a$ for which

$$
\frac{1}{4}<\frac{1}{a+1}+\frac{1}{a+2}+\frac{1}{a+3}<\frac{1}{3}
$$

b) Prove that for any natural number $p \geq 2$ there exist $p$ consecutive natural numbers $a_{1}, a_{2}, \ldots, a_{p}$ such that

$$
\frac{1}{p+1}<\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{p}}<\frac{1}{p}
$$
Solution. a) If $S=\frac{1}{a+1}+\frac{1}{a+2}+\frac{1}{a+3}$, then $\frac{3}{a+3}<S<\frac{3}{a+1}$.

For $a=8$ we get $S=\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$, and $\frac{1}{4}<\frac{1}{12}+\frac{24}{144}=\frac{1}{4}$.

The solutions are $a=8, a=9$ and $a=10$......................................................... $2 \mathbf{p}$

b) For $a_{1}=p^{2}+1, a_{2}=p^{2}+2, \ldots, a_{p}=p^{2}+p$, we have $S=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{p}}=\frac{1}{p^{2}+1}+$ $\frac{1}{p^{2}+2}+\cdots+\frac{1}{p^{2}+p}$ and since $\frac{1}{p^{2}+1}>\frac{1}{p^{2}+2}>\cdots>\frac{1}{p^{2}+p}$, it follows that $\frac{p}{p^{2}+p}<S<\frac{p}{p^{2}+1}$.

Since $\frac{p}{p^{2}+1}<\frac{p}{p^{2}}$, it follows that $\frac{1}{p+1}<S<\frac{1}{p}$