The Cesàro Mean Theorem in the infinite case.

Solution 1:

Since $a_{n}\to\infty$ as $n\to\infty$ then there exists $N$ such that if $n\ge N$ then $a_{n}\ge M>0$. This means at most finitely many terms are negative. By choosing $N$ large enough we may also assume that $\sum_{k=1}^{n}a_{k}>0$ for $n\ge N$. By perhaps choosing $n$ even larger we may assume that $\lvert\frac{\sum_{k=1}^{N}a_{k}}{n}\rvert<1$ So:

$\frac{\sum_{k=1}^{n}a_{k}}{n}=\lvert\frac{\sum_{k=1}^{n}a_{k}}{n}\rvert\ge\frac{\sum_{k=N+1}^{n}a_{k}}{n}-\frac{\sum_{k=1}^{N}a_{k}}{n}>\frac{n-N-1}{n}M-1=(M-1)-\frac{N+1}{n}$

Solution 2:

Given $K>0$, there exists $n_0\in\Bbb N$ such that $a_n\geq 2K$ for $n\geq n_0$. Then, for $n\geq 2n_0$ we have $$\frac 1n\sum_{k=0}^n a_k\geq\frac1n\sum_{k=n_0+1}^n a_k\geq \frac1n (n-n_0)2K=\left(1-\frac{n_0}n\right)2K>K$$