Let $\lbrace x_n\rbrace_{n=1}^{\infty}$ be a sequence of non-negative reals. Is the statement given below is true without any further assumptions? If not, please provide that when these statements mat hold?
If the sequence of arithmetic mean i.e. $\frac{1}{n}\sum_{i=1}^{n} x_i $ converges to $0$ then $x_n$ converges to $0$? What if $0$ is replaced by $c$?
On
Kavi Rama Murthy's example shows that $\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n x_i=c$ doesn't imply $\lim_{n\to\infty} x_n=c$, in the special case $c=\frac{3}{2}$. We can extend to other $c>0$ by scaling the $x_i$. For $c=0$, we can provide the counterexample $x_{n}=\left\{ \begin{array}{rl} 1 & \sqrt{n}\in\mathbb{N}\\ n^{-1} & \sqrt{n}\not\in\mathbb{N} \end{array}\right.$. As with the $c>0$ cases, our choice of $x_n$ has no $n\to\infty$ limit. If however $\lim_{n\to\infty}x_n$ exists and is finite, it's equal to $\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n x_i$.
Take $x_n=1$ if $n$ is even and $x_n=2$ if $n$ is odd and see what happens.