Let's determine the

$$
\lim _{n \rightarrow+\infty} \frac{\sqrt{n^{2}-1^{2}}+\sqrt{n^{2}-2^{2}}+\ldots+\sqrt{n^{2}-(n-1)^{2}}}{n^{2}}
$$

limit.
The $n$-th term of the sequence can be written in the form

$$
\frac{1}{n}\left\{\sqrt{1-\left(\frac{1}{n}\right)^{2}}+\sqrt{1-\left(\frac{2}{n}\right)^{2}}+\ldots+\sqrt{1-\left(\frac{n-1}{n}\right)^{2}}+\sqrt{1-\left(\frac{n}{n}\right)^{2}}\right\}
$$

From this, we recognize that it can be considered as an approximate sum of an integral, specifically the integral of the function $\sqrt{1-x^{2}}$ over the interval $(0,1)$, and the lower approximating sum of rectangles corresponding to the division of the interval into $n$ equal parts. Indeed, the given function is strictly decreasing and monotonic in the interval, so it attains its smallest value at the endpoints of each subinterval

$$
\left(\text{at the points } \frac{1}{n}, \frac{2}{n}, \ldots, \frac{n-1}{n}, \frac{n}{n}\right)
$$

The function is continuous in the interval, hence integrable, positive, and the value of the integral is

$$
\int_{0}^{1} \sqrt{1-x^{2}} d x=\frac{\pi}{4}
$$

since the curve representing the function is the quarter-circle in the first quadrant centered at the origin, so the area under the curve is a quarter of the area of the circle.

And since increasing $n$ beyond any bound makes the division finer beyond any bound, the approximating sum converges to the value of the integral, so the desired limit is $\pi / 4$.

István Gáncs (Győr, Révai M. Gymnasium)