Between which two consecutive whole numbers does the following sum fall?

$$
\frac{1}{\sqrt{17161}}+\frac{1}{\sqrt{17162}}+\frac{1}{\sqrt{17163}}+\ldots+\frac{1}{\sqrt{999999}}+\frac{1}{\sqrt{1000000}}
$$
In the 555th exercise, let's prove 1:

$$
2 \sqrt{n+1}-2 \sqrt{n}<\frac{1}{\sqrt{n}}
$$

For $n=1000000$, we have:

$$
2 \sqrt{1000001}-2 \sqrt{1000000}<\frac{1}{\sqrt{1000000}}
$$

Since $\sqrt{1000001} \approx 1000.0005$ and $\sqrt{1000000} = 1000$, we can approximate:

$$
2 \cdot 1000.0005 - 2 \cdot 1000 < \frac{1}{1000}
$$

which simplifies to:

$$
0.001 < 0.001
$$

This is true, but to make the inequality stronger, we can use the fact that $\sqrt{1000001} > 1000.5$ and $\sqrt{1000000} < 1000.5$ (which can be verified by squaring), and replace these numbers with larger numbers on the right side and smaller numbers on the left side, making the inequalities stronger:

$$
2 \cdot 1000 - 2 \cdot 131 < S < 2 \cdot 1000 - 2 \cdot 130.5
$$

which simplifies to:

$$
1738 < S < 1739
$$

Thus, we have enclosed the desired sum between two consecutive natural numbers.

Zsuzsanna Budai (Budapest, Lorántffy Zs. uti lg. III. o. t.)[^0]


[^0]:    ${ }^{1}$ See the solution in KML XIX. volume 135. p. 1959. November.