Sufficient conditions to guarantee a series is Cesaro summable

Question

Is is true that every nonnegative, bounded series in $R$ is Cesaro summable?

Is there a list of sufficient conditions on series to guarantee that it is Cesaro summable?

Answer 1

In the other case, if you mean the partial sums are non-negative and bounded, no, it is not true that every non-negative bounded series in R is Cesaro summable! This is from somewhere in Hardy's brilliant book 'Divergent Series', look for 'dilution'.

take $a_n = (-1)^n$ and $c_n = \begin{cases}a_k & n=2^k \\ 0 & \text{else} \end{cases} = \begin{cases} (-1)^k & n=2^k \\ 0 & \text{else} \end{cases} $.

We have $$ \sum c_n = 0 + 1 -1 + 0 + 1 + 0 + 0 + 0 -1 + 0 + …$$

By writing $\sigma_k = \sum^k_0 c_n$, we have $$ \sigma_i = 0,1,0,0,1,1,1,1,0,0,...$$ That is, $$\sigma_1 = 1, \sigma_2 = 0, \sigma_4 = 1, \sigma_8 = 0, …$$ and in general for natural $k$, $\sigma_{2^k} = s_k := \begin{cases} 1 & k \text{ even} \\ 0 & k \text{ odd}\end{cases}$. Also, for each $j$ where $2^k \leq j < 2^{k+1}$, we have $\sigma_j = \sigma_{2^k} = s_k$ since the only non-zero term after the $2^k$th term is the $2^{k+1}$th term. That is, $$ \sigma_n = \begin{cases}1 & 2^{k}\leq n<2^{k+1}\text{ for some } k\text{ even} \\ 0 & \text{else}\end{cases} $$ Trying to obtain its $(C,1)$ sum causes problems as follows. Suppose $k=2\kappa+1$ is odd. Then by reasoning similar to the inclusion-exclusion principle from Combinatorics, \begin{align*} \frac{\sum_0^{2^k-1} \sigma_n}{2^k} &=\frac{2^k - 2^{k-1} + \dots + 2 - 1}{2^k} \\ & = \frac{1}{3}\frac{2^{k+1}-1}{2^{k}} \xrightarrow[\kappa\to\infty]{} \frac{2}{3} \end{align*}

but if $k=2\kappa$ is even, then $\sigma_{2^{k-1}},\sigma_{2^{k-1}+1},...\sigma_{2^k-1}$ are all zeros so we instead have \begin{align} \frac{\sum_0^{2^k-1} \sigma_n}{2^k} &=\frac{2^{k-1} - 2^{k-2} + \dots + 2 - 1}{2^k} \\ & = \frac{1}{3}\frac{2^{k}-1}{2^{k}} \xrightarrow[\kappa\to\infty]{} \frac{1}{3} \end{align} So the average of the partial sums of $c_n$ will oscillate between $\frac{1}{3}$ and $\frac{2}{3}$, and the Cesaro sum cannot exist. Coincidentally this series is not Abel summable as well.

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I see your edit now but I will leave this here in case it is helpful to someone else.

Answer 2

If you mean each term of the series is non-negative and bounded this is not true. Since the Cesaro sum of a sequence is the limit of the average of the partial sums, a series of constant terms is not Cesaro summable.

Let $\sigma_n=\frac1n\sum_{k=1}^ns_k$ where $a_n=c>0$ for all $n\in\Bbb N^+$.

$$\sigma_n=\frac cn\sum_{k=1}^n k=\frac{cn(n+1)}{2n}=\frac{c(n+1)}2$$

So, $\sigma_n\to\infty$ as $n\to\infty$.