4. Consider the sequence $\left(x_{n}\right)_{n \geqslant 1}$ of positive real numbers with $\lim _{n \rightarrow \infty} x_{n}=0$. Calculate $\lim _{n \rightarrow \infty} \frac{a_{n}}{n}$, where

$$
a_{n}=\sqrt{2015^{2} x_{1}^{2}+2015 x_{1} x_{2}+x_{2}^{2}}+\sqrt{2015^{2} x_{2}^{2}+2015 x_{2} x_{3}+x_{3}^{2}}+\ldots+\sqrt{2015^{2} x_{n}^{2}+2015 x_{n} x_{1}+x_{1}^{2}}
$$
## Solution.

Since $\left.\sqrt{2015^{2} x^{2}+2015 x y+y^{2}} \leqslant \sqrt{(2015 x+y)^{2}}=2015 x+y, \forall x, y \in \mathbb{R}_{+} \cdot \mathbf{(2 p}\right)$

Thus, we obtain $0 \leqslant a_{n} \leqslant 2016\left(x_{1}+\ldots+x_{n}\right)$, from which $0 \leqslant \frac{a_{n}}{n} \leqslant \frac{2016\left(x_{1}+\ldots+x_{n}\right)}{n}$. (2p)

From $\lim _{n \rightarrow \infty} x_{n}=0$ we get $\lim _{n \rightarrow \infty} \frac{2016\left(x_{1}+\ldots+x_{n}\right)}{n}=0$. (2p)

By the squeeze theorem, it follows that $\lim _{n \rightarrow \infty} \frac{a_{n}}{n}=0$. (1p)