## Task $2 / 89$

Find natural numbers $n$ for which the following holds approximately:

$$
\sum_{i=1}^{n} \frac{1}{\sqrt{i}}=1989
$$
First, we try to find an interval $\left[n_{1} ; n_{2}\right]$ in which approximate solutions to the given equation are certainly located. For every natural number $i \geq 1$, we have

$$
\begin{aligned}
& \frac{1}{\sqrt{i}}=\frac{2}{2 \sqrt{i}}>\frac{2}{\sqrt{i+1}+\sqrt{i}}=2(\sqrt{i+1}-\sqrt{i}) \\
& \frac{1}{\sqrt{i}}=\frac{2}{2 \sqrt{i}}<\frac{2}{\sqrt{i-1}+\sqrt{i}}=2(\sqrt{i}-\sqrt{i-1})
\end{aligned}
$$

From this, it follows that

$$
\sum_{n=1}^{n} 2(\sqrt{i+1}-\sqrt{i})=2(\sqrt{n+1}-1)<\sum_{i=1}^{n} \frac{1}{\sqrt{i}}<\sum_{i=1}^{n} 2(\sqrt{i}-\sqrt{i-1})=2 \sqrt{n}
$$

If we set $2\left(\sqrt{n_{1}+1}-1\right)=1988$ and $2 \sqrt{n_{2}}=1990$, we obtain, after solving these equations, $n_{1}=990024$ and $n_{2}=990025$.

Since these values differ by only 1, both can be considered as approximate values: For $n=n_{1}=990024$, the sum is slightly too small, while for $n=n_{2}=990025$, it is slightly too large.