What is the limit of the following sequence?

$$
a_{n}=\frac{1}{n^{2}}\left[\sin \frac{\pi}{n}+2 \sin \frac{2 \pi}{n}+3 \sin \frac{3 \pi}{n}+\ldots+(n-1) \sin \frac{(n-1) \pi}{n}\right]
$$
We will show that

$$
\lim _{n \rightarrow \infty} a_{n}=\frac{1}{\pi}
$$

First, we simplify \(a_{n}\). From the identity \(\sin \alpha = \sin (\pi - \alpha)\), it follows that

$$
\sin \frac{i \pi}{n} = \sin \frac{(n-i) \pi}{n}
$$

Using this, (1) can be rewritten as:

$$
a_{n} = \frac{1}{n^{2}} \left[ \sin \frac{(n-1) \pi}{n} + 2 \sin \frac{(n-2) \pi}{n} + 3 \sin \frac{(n-3) \pi}{n} + \ldots + (n-1) \sin \frac{\pi}{n} \right]
$$

If we reverse the order of the terms in the brackets and add (1) to (2), we get:

$$
2 a_{n} = \frac{1}{n^{2}} \left[ n \cdot \sin \frac{\pi}{n} + n \cdot \sin \frac{3 \pi}{n} + \ldots + n \cdot \sin \frac{(n-1) \pi}{n} \right]
$$

which simplifies to:

$$
a_{n} = \frac{1}{2 n} \left[ \sin \frac{\pi}{n} + \sin \frac{2 \pi}{n} + \sin \frac{3 \pi}{n} + \ldots + \sin \frac{(n-1) \pi}{n} \right]
$$

The expression in the brackets can be simplified in several ways. The simplest is perhaps the following: Multiply each term in the sum by \(2 \sin \frac{\pi}{2 n}\). Using the identity \(\cos (\alpha - \beta) - \cos (\alpha + \beta) = 2 \sin \alpha \sin \beta\),

$$
\begin{aligned}
& 2 \sin \frac{\pi}{n} \sin \frac{\pi}{2 n} = \cos \frac{\pi}{2 n} - \cos \frac{3 \pi}{2 n} \\
& 2 \sin \frac{2 \pi}{n} \sin \frac{\pi}{2 n} = \cos \frac{3 \pi}{2 n} - \cos \frac{5 \pi}{2 n} \\
& 2 \sin \frac{3 \pi}{n} \sin \frac{\pi}{2 n} = \cos \frac{5 \pi}{2 n} - \cos \frac{7 \pi}{2 n} \\
& \vdots \\
& 2 \sin \frac{(n-1) \pi}{n} \sin \frac{\pi}{2 n} = \cos \frac{(2 n-3) \pi}{2 n} - \cos \frac{(2 n-1) \pi}{2 n}
\end{aligned}
$$

Adding these \(n-1\) equations, we get:

$$
\begin{gathered}
2 \sin \frac{\pi}{2 n} \left[ \sin \frac{\pi}{n} + \sin \frac{2 \pi}{n} + \sin \frac{3 \pi}{n} + \ldots + \sin \frac{(n-1) \pi}{n} \right] = \\
= \cos \frac{\pi}{2 n} - \cos \frac{(2 n-1) \pi}{2 n} = 2 \cos \frac{\pi}{2 n}
\end{gathered}
$$

From this, we obtain the closed form for \(a_{n}\):

$$
a_{n} = \frac{\cos \frac{\pi}{2 n}}{2 n \sin \frac{\pi}{2 n}} = \frac{1}{2 n \tan \frac{\pi}{2 n}}
$$

The sequence \(b_{n} = \cos \frac{\pi}{2 n}\) has a limit as \(n\) approaches infinity:

$$
\lim _{n \rightarrow \infty} b_{n} = \lim _{n \rightarrow \infty} \cos \frac{\pi}{2 n} = 1
$$

Similarly, the sequence \(c_{n} = 2 n \sin \frac{\pi}{2 n}\) also has a limit:

$$
\lim _{n \rightarrow \infty} c_{n} = \lim _{n \rightarrow \infty} 2 n \sin \frac{\pi}{2 n} = \lim _{n \rightarrow \infty} \pi \frac{\sin \frac{\pi}{2 n}}{\frac{\pi}{2 n}} = \pi \lim _{n \rightarrow \infty} \frac{\sin \frac{2 \pi}{n}}{\frac{2 \pi}{n}} = \pi
$$

(We used the fact that \(\frac{\sin x}{x} \rightarrow 1\) as \(x \rightarrow 0\).)

Thus, the sequence \(a_{n}\), which is the ratio of the two sequences, also has a limit, and this limit is the ratio of the limits of the two sequences:

$$
\lim _{n \rightarrow \infty} a_{n} = \frac{\lim _{n \rightarrow \infty} b_{n}}{\lim _{n \rightarrow \infty} c_{n}} = \frac{1}{\pi}
$$

Katalin Drávucz (Szolnok, Verseghy F. Gymnasium, 4th grade)

Remarks. 1. The value of \(\frac{1}{a_{n}}\) is equal to the area of the regular \(2n\)-sided tangential polygon circumscribed around a unit circle. If \(O\) is the center of the circle, \(P_{1}\) and \(P_{2}\) are two adjacent vertices of the regular tangential \(2n\)-sided polygon, then the area of the polygon is \(2n\) times the area of the triangle \(O P_{1} Q\), where \(Q\) is the midpoint of the segment \(P_{1} P_{2}\) (the point of tangency). The length of the segment \(P_{1} Q\) is exactly \(\tan \angle P_{1} O Q = \tan \frac{\pi}{2 n}\), so the area of the tangential \(2n\)-sided polygon is exactly \(2 n \tan \frac{\pi}{2 n}\), which, according to (4), is exactly \(\frac{1}{a_{n}}\). This implies that the sequence \(a_{n}\) is monotonically increasing.

Katalin Drávucz

2. Write \(a_{n}\) in the following form:

$$
\begin{aligned}
a_{n} & = \frac{1}{n} \cdot \frac{1}{n} \sin \frac{\pi}{n} + \frac{1}{n} \cdot \frac{2}{n} \sin \frac{2 \pi}{n} + \frac{1}{n} \cdot \frac{3}{n} \sin \frac{3 \pi}{n} + \ldots + \\
& + \frac{1}{n} \cdot \frac{n-1}{n} \sin \frac{(n-1) \pi}{n} + \frac{1}{n} \cdot \frac{n}{n} \sin \frac{n \pi}{n}
\end{aligned}
$$

(The last term can be freely added to \(a_{n}\) because its value is 0.) From this form, it is immediately clear that \(a_{n}\) is an approximate sum of the integral of the function \(y = x \sin \pi x\) from 0 to 1, specifically the sum corresponding to the uniform partition \(x_{1} = \frac{1}{n}, x_{2} = \frac{2}{n}, x_{3} = \frac{3}{n}, \ldots, x_{n-1} = \frac{n-1}{n}, x_{n} = \frac{n}{n} = 1\). Therefore,

$$
\lim _{a_{n} \rightarrow \infty} a_{n} = \int_{0}^{1} x \sin \pi x \, dx = \frac{1}{\pi}
$$

Gábor Megyesi (Szeged, Ságvári E. Practical Gymnasium, 1st grade)

3. Similarly, from (3), it is immediately clear that \(2 a_{n}\) is the approximate sum of the integral of the function \(y = \sin \pi x\) from 0 to 1, corresponding to the same partition. Therefore,

$$
\lim _{n \rightarrow \infty} a_{n} = \int_{0}^{1} \sin \pi x \, dx = \frac{1}{\pi}
$$

Gábor Hetyei (Pécs, Leövey K. Gymnasium, 3rd grade)

4. Let \(z\) be the complex number \(\left(\cos \frac{\pi}{n} + i \sin \frac{\pi}{n}\right)\). Then, on the right-hand side of (3), the terms in the brackets are the imaginary parts of the complex numbers \(z, z^{2}, z^{3}, \ldots, z^{n-1}\). Consequently, \(a_{n}\) is the \(\frac{1}{2n}\)-th part of the imaginary part of the sum \(z + z^{2} + z^{3} + \ldots + z^{n-1}\). Using the formula for the sum of a geometric series and the fact that \(z^{n} = -1\), it follows that \(a_{n}\) is the \(\frac{1}{2n}\)-th part of the imaginary part of the complex number \(\frac{1+z}{1-z} = \frac{2}{1-z} - 1\). It is easy to verify that \(\frac{2}{1-z} = 1 + i \cot \frac{\pi}{2 n}\), from which (4) immediately follows.