Cesaro summable series

Question

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.

Answer 1

So it will save some notation to just rephrase this problem in terms of sequences (as you started doing). We say $(a_n)$ cesaro-converges to $\ell$ if the sequence of its running averages $\sigma_n$ converges to $\ell$. We are trying to prove cesaro-convergence is implied by ordinary convergence.

Let $\epsilon >0$. Let $N$ be such that all terms $a_k$ with $k > N$ satisfy $$ |a_k-\ell| < \epsilon/3. $$

Now you can look at $\sigma_M$ for $M$ large enough as $$ \sigma_M = \frac{a_1+\cdots+a_N}{M} + \frac{a_{N+1} + \cdots + a_M}{M}. $$

The second summand can be viewed as: $$ \frac{M-N}{M}\cdot \frac{a_{N+1} + \cdots + a_M}{M-N}, $$ the product of something tending to 1 and something which is $\epsilon/3$-close to $\ell$, by choice of $N$. So for $M$ large enough, it will be $\epsilon/2$-close to $\ell$.

The first summand tends to 0. So by choosing $M$ large enough, you can get the entire expression $\epsilon$-close to $\ell$.