Let $n$ be a positive integer. Determine the first digit after the decimal point of the number

$$
\sum_{k=1}^{n} \frac{\sqrt{k(k+1)}}{n}
$$
Solution. For every positive integer $k$,

$$
(k+0.4)^{2}=k^{2}+0.8 k+0.16 \leq k^{2}+0.96 k<k^{2}+k<k^{2}+k+0.25=(k+0.5)^{2}
$$

thus $k+0.4<\sqrt{k(k+1)}<k+0.5$. Therefore,

$$
\sum_{k=1}^{n} \frac{k+0.4}{n}<\sum_{k=1}^{n} \frac{\sqrt{k(k+1)}}{n}<\sum_{k=1}^{n} \frac{k+0.5}{n}
$$

Transform the two extreme sums:

$$
\begin{aligned}
& \sum_{k=1}^{n} \frac{k+0.4}{n}=\frac{\frac{(1+n) \cdot n}{2}+0.4 \cdot n}{n}=\frac{n}{2}+0.5+0.4=\frac{n}{2}+0.9 \\
& \sum_{k=1}^{n} \frac{k+0.5}{n}=\frac{\frac{(1+n) \cdot n}{2}+0.5 \cdot n}{n}=\frac{n}{2}+0.5+0.5=\frac{n}{2}+1
\end{aligned}
$$

Thus,

$$
\frac{n}{2}+0.9<\sum_{k=1}^{n} \frac{\sqrt{k(k+1)}}{n}<\frac{n}{2}+1
$$

If $n$ is even, then $\frac{n}{2}$ is an integer, so the sum falls between $\frac{n}{2}+0.9$ and $\frac{n}{2}+1$, and thus the first digit after the decimal point is 9.

If $n$ is odd, then $\frac{n}{2}-0.5$ is an integer, and the sum falls between

$$
\begin{aligned}
\left(\frac{n}{2}-0.5\right)+0.5+0.9 & =\left(\frac{n}{2}-0.5\right)+1.4 \quad \text { and } \\
\left(\frac{n}{2}-0.5\right)+0.5+1 & =\left(\frac{n}{2}-0.5\right)+1.5
\end{aligned}
$$

thus the questioned digit is 4.