I am trying to extend the Cesaro Means result here: http://www.ee.columbia.edu/~vittorio/CesaroMeans.pdf
The author there proved: $\lim_{n\to\infty} a_n = a \implies \lim_{n\to\infty} 1/n \sum_{i=1}^n a_i = a$.
I want to extend this to: $\lim_{n\to\infty} a^{(n)}_n = a \implies \lim_{n\to\infty} 1/n \sum_{i=1}^n a^{(n)}_i = a$.
My attempt so far:
WLOG assume $a=0$. $$ |1/n \sum_{i=1}^n a^{(n)}_i| \leq 1/n \sum_{i=1}^{n-1} |a^{(n)}_i| + a^{(n)}_n/n $$ $$ \implies |1/n \sum_{i=1}^n a^{(n)}_i| \leq \frac{(n-1) \max_{i=1,\dots,n-1}|a^{(n)}_i| + a^{(n)}_n}{n} $$ So I'm stuck here. This is not an exercise in a book so I'm not sure if my claim is even correct. If it is not true, what are the minimal assumptions I need to place on $a^n_n$ for this to be true.
It can't be generally true. Try a simple example like $a_n = 2^{1/n}$ where $a_n^n \to 2$.
Do you think $\frac{1}{n} \sum_{k=1}^n 2^{n/k} \to 2$?
At least $n/4$ terms are bigger than $2^4/n$, $n/8$ terms are bigger than $2^8/n$, etc.