1
$\begingroup$

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 $-\infty$?

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

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 $x_i=(−1)i.$"

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.