prove $\sum |{x_n}|$ converges, then $\sum {x_n}$ converges

Question

I need to prove that $\sum |{x_n}|$ converges, then $\sum {x_n}$ converges too. However, isn't this false and the converse is true instead?

Answer 1

HINT

Note that $x_n \le |x_n|$, and then $x_n \ge -|x_n|$,

hence $-\sum |x_n|\le\sum x_n \le \sum |x_n|$.

Answer 2

Suppose that $\sum_{n=1}^\infty |x_n| < \infty$, and consider the sequence $S_n = \sum_{i=1}^\infty x_i$ of partial sums. Then we will show $(S_n)_{n=1}^\infty$ is a Cauchy sequence and therefore has a limit. Note that for $m < n$, we have $$ |S_n - S_m| = \left| \sum_{i=m+1}^n x_n \right| \le \sum_{i=m+1}^n |x_i| \le \sum_{i=m+1}^\infty |x_i|. $$ However, as $m \to \infty$, the right hand side approaches a limit of zero. From this, the Cauchy property of $(S_n)_{n=1}^\infty$ will follow with a little work.

Answer 3

The other answer I presented is a very general solution, and is one that you'll probably see again if you go on to study functional analysis, and Banach spaces in particular.

On the other hand, for the case of real sequences (and a similar argument will also work for complex sequences), there is another solution which you might find more intuitive. The basic idea is that $\sum_{n=1}^\infty |a_n|$ converging implies that the positive terms and the negative terms both have converging sums by the comparison theorem for series of nonnegative sequences.

To make the argument precise, given $x \in \mathbb{R}$, we will define $$ x^+ := \begin{cases} x, & x \ge 0;\\ 0, & x < 0, \end{cases} \mathrm{~and~} x^- := \begin{cases} 0, & x \ge 0;\\ -x, & x < 0. \end{cases}$$ Then I will leave it as an exercise to show that for any $x \in \mathbb{R}$, we have $0 \le x^+, x^- \le |x|$; $x = x^+ - x^-$; and $|x| = x^+ + x^-$.

Now, by the comparison theorem (comparing to $\sum_{n=1}^\infty |x_n|$), we get that $\sum_{n=1}^\infty x_n^+$ and $\sum_{n=1}^\infty x_n^-$ are both convergent series. Therefore, $\sum_{n=1}^\infty x_n = \sum_{n=1}^\infty (x_n^+ - x_n^-)$ is also a convergent series.

Answer 4

$S_n=\sum_{k=1}^{n}x_k.$

$T_n = \sum_{k=1}^{n}|x_k|.$

$T_n$ converges, implies $T_n$ is a Cauchy sequence.

$\epsilon >0$ given, there is a $n_0$, s.t.

for $m \ge n \ge n_0$:

$|T_m-T_n|= |\sum_{k=n+1}^{m}|x_k|| <\epsilon$;

We have:

$|S_m-S_n| = |\sum_{k=n+1}^{m}x_k| \le $

$|\sum_{k=n+1}^{m}|x_k||= |T_m-T_n| \lt \epsilon$ .

$S_n$ is Cauchy, hence convergent.