Convergence of partial sums of real sequences

Question

For all $i\in\mathbb{N}$, let $(a_{i,n})_{n\in\mathbb{N}}$ be a real sequence that tends to $0$ for $n\rightarrow\infty$. It holds also that $|a_{i,n}|\leq1$ for all $i,n\in\mathbb{N}$. Is it possible to show that \begin{align*} c_n:=\frac{1}{n}\sum_{i=1}^{n}a_{i,n}\xrightarrow{n\rightarrow\infty}0?\end{align*} Thanks!

Answer 1

Counter-example: $$a_{i,n}=\left\{\begin{array}{ll} 1&n< 2i\\0 & n\ge 2i\end{array}\right.\ .$$ It is easy to see that $\lim\limits_{n\to\infty}c_n=\frac{1}{2}$.

Answer 2

• $ a_{i,n} ≤ 1 $
• $ ∑a_{i,n} ≤ n $
• $ (∑a_{i,n}) /n ≤ 1 $
• $ lim(∑a_{i,n}/n) \rightarrow0$

finaly $C_n \rightarrow0$

Answer 3

I think the answer depends on the sequences themselves. Here is what I have thought.

$\forall\ i\in \mathbb{N},\ \forall \epsilon>0,\ \exists N_i\in \mathbb{Z}^+$ s.t. $$|a_{i,n}|<\epsilon$$ for all $n\geq N_i$. Then let $N=\max_{i\in \mathbb{N}}N_i$. Now $\mathbf{IF}\ N<\infty$ then we see that $$|\frac{1}{n}\sum_{i=1}^na_{i,n}|\leq\frac{1}{n}\sum_{i=1}^n |a_{i,n}|<\epsilon$$ for all $n\geq N$, and consequently the sequence $c_n$ converges to $0$. But it does not seem obvious that the $\mathbf{IF}$ always holds.