Is the Cesàro summation of a sequence divergent to infinity divergent? [duplicate]

More specifically the question is: If I have a sequence $(u_n)_{n\in\mathbb{N}} \subset \mathbb{R}$ that diverges to infinity. Then it's Cesàro summation sequence $(s_n)_{n\in\mathbb{N}}=(\frac{1}{n+1}\sum_{k=0}^n u_k)_{n\in\mathbb{N}}$ is also divergent to infinity.

This seems true but I can't prove it.

Solution 1:

Yes. Let $a\in\Bbb R$. As $u_n\to +\infty$, there exists $N$ such that $u_n>a+1$ for all $n>N$. Then $(n+1)s_n-(N+1)s_N>(n-N)(a+1)$ for $n>N$ and so $$s_n>\frac{n-N}{n+1}\cdot (a+1) +\frac{N+1}{n+1}s_N=a+1+\frac{N+1}{n+1}(s_N-a-1).$$ For $n>\max\{N,(N+1)(s_N-a-1)\}$, this means $s_n>a$.