## Task 6 - 341236

Determine for each odd natural number $n \geq 3$ the number

$$
\left[\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+\ldots+\frac{1}{\sqrt{n^{2}-4}+\sqrt{n^{2}-3}}+\frac{1}{\sqrt{n^{2}-2}+\sqrt{n^{2}-1}}\right]
$$

Hint: If $z$ is a real number, then [ $z]$ denotes the integer $g=[z]$, for which $g \leq z<g+1$ holds.
We first expand each fraction as follows:

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

Then let the bracketed sum be

$$
\begin{gathered}
s=-\sqrt{1}+\sqrt{2}-\sqrt{3}+\sqrt{4}-+\ldots-\sqrt{n^{2}-2}+\sqrt{n^{2}-1} \\
s=\sum_{k=1}^{\frac{1}{2}\left(n^{2}-1\right)}(\sqrt{2 k}-\sqrt{2 k-1})
\end{gathered}
$$

We now use the following inequality:

$$
\sqrt{2 k}-\sqrt{2 k-1}\sqrt{n^{2}-1}-1-\frac{1}{\sqrt{2}} \sum_{k=1}^{\frac{1}{2}\left(n^{2}-3\right)}\left(\sqrt{k+\frac{1}{2}}-\sqrt{k-\frac{1}{2}}\right)
$$

This is also a telescoping sum, and therefore:

$$
\begin{gathered}
s>\sqrt{n^{2}-1}-1-\frac{1}{\sqrt{2}}\left(\sqrt{\frac{1}{2}\left(n^{2}-3\right)+\frac{1}{2}}-\sqrt{1-\frac{1}{2}}\right) \\
s>\sqrt{n^{2}-1}-1-\frac{1}{2}\left(\sqrt{n^{2}-2}-1\right) \\
s>\sqrt{n^{2}-1}-\frac{1}{2} \sqrt{n^{2}-2}-\frac{1}{2}
\end{gathered}
$$

We now show that

$$
\begin{gathered}
\sqrt{n^{2}-1}-\frac{1}{2} \sqrt{n^{2}-2}>\frac{1}{2} n \\
2 \sqrt{n^{2}-1}>n+\sqrt{n^{2}-2} \\
4 n^{2}-4>n^{2}+n^{2}-2+2 n \sqrt{n^{2}-2} \\
n^{2}-1>n \sqrt{n^{2}-2} \\
n^{4}-2 n^{2}+1>n^{4}-2 n^{2}
\end{gathered}
$$

This is always true, so we can substitute in (5):

$$
s>\frac{1}{2} n-\frac{1}{2}=\frac{1}{2}(n-1)
$$

Together with equation (3), this gives

$$
\frac{1}{2}(n-1)<s<\frac{1}{2} n
$$

Since \( n \) is supposed to be an odd number, it is clear that

$$
g=\frac{n-1}{2}
$$