# Convergence $\overset{?}{\Rightarrow}$ absolute convergence (of a series)

What is wrong with the proof below?

Suppose $\displaystyle\sum_{n=1}^\infty a_n$ converges. Then it converges absolutely.

Proof. $\quad \forall_{\epsilon>0}\ \exists_{N}\ m\geq n \geq N \Rightarrow \left\lvert x_m - x_n \right\rvert < \epsilon \quad$ (Cauchy property)

But $\left\lvert \lvert x_m \rvert - \lvert x_n \rvert \right\rvert \leq \left\lvert x_m - x_n \right\rvert \Rightarrow \left\lvert \lvert x_m \rvert - \lvert x_n \rvert \right\rvert < \epsilon \qquad$ q.e.d.

• First you have to know the difference between sequence and series. – corindo Sep 21 '15 at 7:54

I assume that $x_n=a_1+\dots+a_n$. To prove that $\sum_{n=1}^{\infty}|a_n|$ converges, you have to prove (if you wish) that for all $\epsilon>0$ $\exists N$ such that for all $m\geq n\geq N$ $|a_n|+\dots +|a_m|<\epsilon$. You only proved that $||a_0+\dots +a_m|-|a_0+\dots +a_n||<\epsilon$.
As the error has been pointed out, I provide a counterexample: The series $\sum_{n \geq 1} (-1)^{n-1}\frac{1}{n}$ converges by Leibniz's rule, but $\sum_{n\geq 1}\frac{1}{n}$ diverges.