I've just seen this:
$\displaystyle \{x_i\}_{i=1}^\infty \ \sum _i ^\infty |x_i| < + \infty \rightarrow lim_{+\infty} x_i = 0$
why is this true?
Let $\displaystyle \sum_{i\,=\,1}^\infty|x_i|=L.$ Then given $\displaystyle \epsilon>0,$ $\displaystyle \exists\,N\in\mathbb N$ such that $\displaystyle n\geqslant N\ \Rightarrow\ \left|\sum_{i\,=\,1}^n|x_i|-L\right|<\frac\epsilon2.$ Hence, $\displaystyle \forall\,n\geqslant N+1,$
$\displaystyle |x_n|$
$\displaystyle =\ ||x_n||$
$\displaystyle =\ \left|\sum_{i\,=\,1}^n|x_i|-\sum_{i\,=\,1}^{n-1}|x_i|\right|$
$\displaystyle =\ \left|\left(\sum_{i\,=\,1}^n|x_i|-L\right)-\left(\sum_{i\,=\,1}^{n-1}|x_i|-L\right)\right|$
$\displaystyle \leqslant\ \left|\sum_{i\,=\,1}^n|x_i|-L\right|+\left|\sum_{i\,=\,1}^{n-1}|x_i|-L\right|$
$\displaystyle <\ \frac\epsilon2+\frac\epsilon2$
$\displaystyle =\ \epsilon$
To say that $\displaystyle \sum _i ^\infty |x_i| < + \infty$ means $\displaystyle S_n = \sum\limits_{k = 1}^n {\left| {x_k } \right|} \, \Rightarrow \,\left( {\exists L} \right)\left[ {\left( {S_n } \right) \to L} \right]$.
So the sequence $\displaystyle S_n$ is a Cauchy sequence.
That means we can make $\displaystyle \left| {x_n } \right| = \left| {S_n - S_{n - 1} } \right| < \varepsilon $.
That is sufficient to prove the property: $\displaystyle \left| {x_n } \right| \to 0$
No, but close: since the sequence (of partial sums of the series) is Cauchy then it has a limit, but since the limit of $\displaystyle S_n$ equals the limit of $\displaystyle S_{n-1}$ and since, as I wrote in the first message, $\displaystyle |x_n|=S_n-S_{n-1}$, then...
Be careful: $\displaystyle S=\lim_{n\rightarrow \infty}S_n$ can NOT be zero unless $\displaystyle x_i=0\,\,\forall\,i$ !
Tonio