## Problem 2

Consider the sequence $\left(x_{n}\right) \mathrm{n} \geq 1$ of positive real numbers with $\lim _{n \rightarrow \infty} \mathrm{x}_{\mathrm{n}}=0$.

Calculate:

$\lim _{n \rightarrow \infty} \frac{\sqrt{x_{1}^{2}-x_{1} x_{2}+x_{2}^{2}}+\sqrt{x_{2}^{2}-x_{2} x_{3}+x_{3}^{2}}+\cdots+\sqrt{x_{n}^{2}-x_{n} x_{1}+x_{1}^{2}}}{n}$

GMB 4/2012
## Problem 2

Consider $a_{n}=\sqrt{x_{1}^{2}-x_{1} x_{2}+x_{2}^{2}}+\sqrt{x_{2}^{2}-x_{2} x_{3}+x_{3}^{2}}+\cdots+\sqrt{x_{n}^{2}-x_{n} x_{1}+x_{1}^{2}}, \mathrm{n} \geq 3$

$$
\begin{aligned}
& \text{Since} \sqrt{x^{2}-x y+y^{2}} \leq \sqrt{(x+y)^{2}}=\mathrm{x}+\mathrm{y}, \forall \mathrm{x}, \mathrm{y} \geq 0 \Rightarrow \\
& \Rightarrow 0 \leq a_{n} \leq x_{1}+x_{2}+x_{2}+x_{3}+\cdots+x_{n}+x_{1}=2\left(x_{1}+x_{2}+\cdots+x_{n}\right)
\end{aligned}
$$

Then $0 \leq \frac{a_{n}}{n} \leq \frac{2\left(x_{1}+x_{2}+\ldots+x_{n}\right)}{n}$

Since $\lim _{n \rightarrow \infty} x_{n}=0 \xrightarrow{\text{ by Stolz }} \lim _{n \rightarrow \infty} \frac{x_{1}+x_{2}+\cdots+x_{n}}{n}=0$ (1p) $=\lim _{n \rightarrow \infty} \frac{a_{n}}{n}=0$