Question about Cesàro summation Consider $$S_n = \sum_{i=0}^n a_i$$
and its Cesàro sums, defined as
$$ C = \lim_{n \to \infty} \frac1n\sum_{k=0}^n S_k$$
Is it always true that
$$ C = \lim_{n \to \infty} \frac1{L(n)}\sum_{k= n - L(n)}^n S_k$$
where $L(n)$ is any strictly increasing function such that $ 2 < L(n) < \ln(n)$ for every $n$?

tl;dr: no.

Define for convenience $L(n)$; by assumption, we have $L(n)=o(n)$ as $n\to\infty$.


*

*Note that if $L(n)$ is bounded, then this is clearly false as the few missing constant terms do not  really matter: you cannot hope that $\sum_{k=0}^n S_k$ be bounded in general (that is, there are Cesàro summable sums which are not convergent in the usual summation sense). For instance, take $a_n=(-1)^n$.

*Assuming now $L(n)\xrightarrow[n\to\infty]{} \infty$, this looks like you are wondering about some specific converse of the Stolz–Cesàro theorem.
However, as Did's comment above shows, this is still false even with this assumption: I reproduced this comment below:

This cannot hold. Try $a_n=1$ if $n=3^k-k$ and $a_n=-1$ if $a_n=3^k$, for some $k\geqslant1$, thus the first terms of the sequence $(a_n)_{n\geqslant0}$ are $0|0|1|-1|0|0|0|1|0|-1|0$ and $S_n=1$ if $3^k-k\leqslant n<3^k$ for some $k\geqslant1$ while $S_n=0$ for every other $n$. In particular, $(S_n)$ is Cesàro-summable with $C=0$ but, for
  $$L(n)=\lfloor\log_3(n)\rfloor$$
  the ratios
  $$\frac1{L(n)}\sum_{k=n-L(n)}^nS_k$$
  fluctuate between 0 and 1.

