Density argument This is a problem that I met when studying density things:

Let $\{ a_n \}$ be a real-valued sequence. Then the following things are equivalent:
(1) $\lim_{n \rightarrow \infty } \frac{1}{n}\sum_{i=1}^{n}a_n=0$
(2) There exists some $J \subset \mathbb{N}$ with natural density $0$, such that $\lim_{n \rightarrow \infty, n \notin J} a_n=0$

I'm a beginner on Density arguments, so can somebody help?
 A: The statement is wrong. If $a_n=(-1)^{n}$ then 1) holds but 2) doesn't.

However we have the following:
Let $\{a_{n}\}⊂ℝ$ be bounded. Then the following are equivalent:
a) $(1/n)\sum\limits_{k=1}^{n}|a_{k}|→0$
b) there exists $A\subset \mathbb N$ such that $\lim_{n\in A,n \to \infty} a_{n}=0$ and $card(A∩\{1,2,...,n\})/n)→0$ as n→∞
Here is a stronger result:
Let $\{a_{n}\}\subset 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ be bounded and $1\leq p<\infty $. Then the following are equivalent:
a) $\frac{1}{n}\sum\limits_{k=1}^{n}\left\vert a_{k}\right\vert \rightarrow
0$
b) there exists $A\subset 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
$ such that $\underset{n\notin A,n\rightarrow \infty }{\lim }a_{n}=0$ and $%
\frac{\#\{A\cap \{1,2,...,n\})}{n}\rightarrow 0$ as $n\rightarrow \infty $
c) $\frac{1}{n}\sum\limits_{k=1}^{n}\left\vert a_{k}\right\vert
^{p}\rightarrow 0$
It suffices to show that a) and b) are equivalent. Suppose $\frac{1}{n}%
\sum\limits_{k=1}^{n}\left\vert a_{k}\right\vert \rightarrow 0$. For $%
k=1,2,...$ let $I_{k}=\{n\geq 1:\left\vert a_{n}\right\vert \geq \frac{1}{k}%
\}$. Claim: $\frac{\#\{I_{k}\cap \lbrack 1,n]\}}{n}\rightarrow 0$ as $%
n\rightarrow \infty $ for each $k$. Indeed, this follows from the inequality
$\frac{1}{n}\sum\limits_{j=1}^{n}\left\vert a_{j}\right\vert \geq \frac{%
\#\{I_{k}\cap \lbrack 1,n]\}}{nk}$. There exist integers $n_{0}<n_{1}<...$
such that $n\geq n_{k}$ implies $\frac{\#\{I_{k+1}\cap \lbrack 1,n]\}}{n}<%
\frac{1}{k+1}$. Let $I=\bigcup\limits_{k=0}^{\infty }\{I_{k+1}\cap \lbrack
n_{k},n_{k+1})\}$. Let $n_{k}\leq n<n_{k+1}$. Then $I\cap \lbrack
1,n]\subset \{I_{k}\cap \lbrack 1,n_{k}]\}\cup \{I_{k+1}\cap \lbrack 1,n]\}$%
. Hence $\frac{\#\{I\cap \lbrack 1,n]\}}{n}\leq \frac{\#\{I_{k}\cap \lbrack
1,n_{k}]\}}{n}+\frac{\#\{I_{k+1}\cap \lbrack 1,n]\}}{n}<\frac{n_{k}}{n}\frac{%
1}{k}+\frac{1}{k+1}\leq \frac{1}{k}+\frac{1}{k+1}$ if $n\geq n_{k}$. We have
proved that $\frac{\#\{I\cap \lbrack 0,n)\}}{n}\rightarrow 0$ as $%
n\rightarrow \infty $. If $n>n_{k}$ and $n\notin I$ then $n\notin I_{k+1}$
( for, otherwise, there exists $\rho \geq k$ such that $n_{\rho }\leq
n<n_{\rho +1}$ and $n\in I_{k+1}\subset I_{\rho +1}$ so $n\in I_{\rho
+1}\cap \lbrack n_{\rho },n_{\rho +1)}\subset I$ which is a contradiction).
Thus $\left\vert a_{n}\right\vert <\frac{1}{k+1}$ for $n>n_{k}$, $n\notin I$
completing the proof of a) implies b). For the converse part let $\left\vert
a_{n}\right\vert \leq C$ and let $\epsilon >0$. There exists $n_{\epsilon }$
such that $\left\vert a_{n}\right\vert <\epsilon $ if $n>n_{\epsilon }$ and $%
n\notin I$. Also there exists $m_{\epsilon }$ such that $\frac{\#\{I\cap
\lbrack 0,n)\}}{n}<\epsilon $ if $n>m_{\epsilon }$. For $n>\max
\{n_{\epsilon },m_{\epsilon }\}$ we have $\frac{1}{n}\sum%
\limits_{k=0}^{n-1}\left\vert a_{k}\right\vert <\epsilon +\epsilon C$.
