Suppose you have a sum $\sum a_n $ which is Cesaro-summable. I.E. the average of the partial sums of
$$ a_1 + a_2 + a_3 + a_4 … $$
is well defined and converges. Now suppose you throw a bunch of zeros in there but not too many, every term $a_n$ has finitely many zero's before it, so for example you could transform the series to the following (among many other options which are much more zero dense)
$$ a_1 + 0 + a_2 + 0 + a_3 + 0 + a_4 + … $$
$$ a_1 + 0 + a_2 + 0 + 0 + a_3 + 0 + 0 + 0 + a_4 + … $$
It seems that the Cesaro sums of these series still converge and that by throwing lots of 0 you can slow down the convergence arbitrarily high but you can never prevent it from converging, as long as your sprinkling of zeros still allows every summand $a_i$ to appear after a FINITE number of preceding terms.
How to prove this? Or: to find a counter example to it. I.E. a sequence of numbers $n_1, n_2, n_3…$ such that if you put $n_1$ $0$'s after $a_1$ and $n_2$ $0$'s after $a_2$ etc… that the cesaro sum of this sequence will not be defined.
Best Answer
You have a sequence of partial sums $\{S_n\}_{n=1}^{\infty}$.
By adding an arbitrary but finite number of zeros after each term, you have new partial sums $\{\tilde{S}_n\}_{n=1}^{\infty}$. The new partial sum sequence is the same as the old except each term $S_n$ is repeated however many (finite) times you like. For example $$\{\tilde{S}_n\} = \{S_1, S_2, S_2, S_3, S_3, S_3, S_3, S_4, ...\}$$
Fix $c \in \mathbb{R}$. If $S_n\rightarrow c$ then $\tilde{S}_n\rightarrow c$.
On the other hand, suppose we only know: $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^n S_i = c$$ Then $\frac{1}{n}\sum_{i=1}^n S_i$ may not converge. Try $$a_i = (-1)^{i+1} \quad \forall i \in \{1, 2, 3, ...\}$$ Then $S_i=1$ if $i$ is odd and $S_i=0$ if $i$ is even, and $$ \lim_{n\rightarrow\infty} \frac{1}{n}\sum_{i=1}^n S_i = \frac{1}{2}$$ However, we can just repeat terms to make $\frac{1}{n}\sum_{i=1}^n \tilde{S}_i$ converge to something other than $\frac{1}{2}$, or not converge at all. For example \begin{align} &\{S_n\} = \{1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...\}\\ &\{\tilde{S}_n\} = \{1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...\} \end{align} We can make $\{\tilde{S}_n\}$ any arbitrary binary-valued sequence that starts with 1 and contains an infinite number of 0s and an infinite number of 1s.