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.
On
Let $M$ be any positive real number. Let $s_n = \sum_{k=0}^{n}u_k$. Since $s_n \to \infty$, there is some $N$ such that $s_n > M$ for all $n > N$. Consequently, if $n > N$ we have
$$\sum_{k=0}^{n}s_k = \sum_{k=0}^{N} s_k + \sum_{k=N+1}^{n} s_k > \sum_{k=0}^{N}s_k + (n-N)M$$ and so $$\frac{1}{n+1}\sum_{k=0}^{n}s_k > \frac{1}{n+1}\sum_{k=0}^{N}s_k + \frac{n-N}{n+1}M$$ As $n \to \infty$, the first term on the right hand side converges to zero (since $N$ is fixed), and the second term converges to $M$. Therefore $$\liminf_{n \to \infty}\frac{1}{n+1}\sum_{k=0}^{n}s_k \geq M$$ Since this is true for arbitrary $M$, we conclude that $$\liminf_{n \to \infty}\frac{1}{n+1}\sum_{k=0}^{n}s_k = \infty$$ In other words, the Cesaro sum diverges to infinity.
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 $s_n>a$.