A question emerging from an exercise in Ok, E. A. (2007). Real Analysis with Economic Applications. Princeton University Press.

The exercise consists in showing that if $\sum_{i=1}^\infty x_i$ converges, then

$|\sum_{i=k}^\infty x_i| \leq \sum_{i=k}^\infty |x_i|$.

I do not get why the condition $\sum_{i=1}^\infty x_i$ is necessary. Why can't we say in general that

$|\sum_{i=k}^\infty x_i| = |~lim_{n\rightarrow \infty}~ x_k + x_{k+1} + \dots + x_{k+n}|$

$~~~~~~~~~~~~~~= ~lim_{n\rightarrow \infty}~ |~x_k + x_{k+1} + \dots + x_{k+n}|$

$~~~~~~~~~~~~~~\leq ~lim_{n\rightarrow \infty}~ |x_k| + |x_{k+1}| + \dots + |x_{k+n}|,$ by the triangular inequality

$~~~~~~~~~~~~~~ = ~\sum_{i=k}^\infty |x_i|$

Is it that $|\sum_{i=k}^\infty x_i|$ is not well-defined when $\sum_{i=k}^\infty x_i \pm \infty$, in which case the convergence of $\sum_{i=1}^\infty x_i$ would be a way to guarantee that $\sum_{i=k}^\infty x_i \neq \pm \infty$? The standard definition of the absolute value function is over reals, and not the extended reals but it seems like it is not a fundamental problem to extend it to $\infty$ and $-\infty$?

  • 2
    $\begingroup$ There are worse ways for $\sum_ix_i$ to be undefined: consider $x_i=(-1)^i$. $\endgroup$ Jun 1 '13 at 16:03
  • $\begingroup$ Sure, thank you. The author warned about this a page before :-S $\endgroup$ Jun 1 '13 at 16:12

Just so that I can mark the question as answered in two days:

As Brian noted "There are worse ways for $\sum_{i=1}^{\infty}x_i$ to be undefined: consider $x_i=(−1)i.$"


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