Theorem 3.22 from baby Rudin $\sum a_n$ converges if and only if for every $\varepsilon >0$ there is an integer $N$ such that $$\left|\sum_{k=n}^{m}a_k\right|\leqslant \varepsilon$$ if $m\geqslant n\geqslant N$. 
In particular, by taking $m=n$ above inequality becomes $$|a_n|\leqslant \varepsilon \quad(n\geqslant N).$$
In other words: If $\sum a_n$ converges, then $\lim_{n\to \infty}a_n=0$
The condition $a_n\to 0$ is not, however, sufficient to ensure convergence of $\sum a_n$. For instance. the series $\sum \frac{1}{n}$ diverges.
Reading all this I have one question. 
If for every $\varepsilon >0$ there is an integer $N$ such that $\left|\sum_{k=n}^{m}a_k\right|\leqslant \varepsilon$ if $m\geqslant n\geqslant N$ then $\sum a_n$ converges. 
If we put here $m=n$ why can not conclude that $\sum a_n$ converges? Where is the mistake?

The mistake is that
$$
\sum_{k=m}^m a_k = a_m
$$
and you are saying nothing about the series $\sum_k a_k$. You are only looking at the single term $a_m$.
