A series $\sum_{k=0}^{∞}a_k$ is said to be Cesaro summable to an $L\in R$ if and only if $\sigma_n = \sum_{k=0}^{n-1}(1 - \frac{k}{n})a_k$ converges to $L$ as $n$ → $∞$.
Let $s_n = \sum_{k=0}^{n-1}a_k$ be the partial sums of $\sum_{k=0}^{∞}a_k$. And let $\sigma_{n} = \frac{s_1 + \cdots+ s_n}{n}$ for every natural $n$.
Exercise: Prove that if ${a_k}$ is real sequence and $\sum_{k=0}^{∞}a_k = L$ converges, then $\sum_{k=0}^{∞}a_k$ is Cesaro summable to $L$.
Attempt in proof: Suppose ${a_k}$ is real sequence and $\sum_{k=0}^{∞}a_k = L$ converges. Then by definition, $\sum_{k=0}^{∞}a_k = L$ converges if and only if its sequence of partial sums ${s_n}$ converges to $L\in R$. That is, for every $\epsilon > 0$there is an $N \in N$ such that $n \geq N$ implies $|s_n - L| < \epsilon$.
Or we could show if $|\sigma_{n} - L| < \epsilon $ then $\sum_{k=0}^{∞}a_k$ is Cesaro summable. Then $|\sigma_{n} - L| = | \frac{s_1 + \cdots+ s_n}{n} - L| = |\sum_{k=0}^{n-1}(1 - \frac{k}{n})a_k - L|$
Can someone please help me ? I don't know how to simplify, to see if the expression will converge. I would really appreciate it.
Thank you.