5
$\begingroup$

If $\sum\limits_{n=0}^{\infty} a_n\ $ is a conditionally convergent series, then is the sum of the average of consecutive terms necessarily convergent?

In other words, is $\sum\limits_{n=0}^{\infty} |a_n+a_{n+1}|\ $ convergent?

For the most standard example, the alternating harmonic series $\sum\limits_{n=0}^\infty {(-1)^n\over n+1},$ we get that $\sum\limits_{n=0}^{\infty} |a_n+a_{n+1}|\ $ is half the sum of the reciprocal triangular numbers, which does converge.

But does every conditionally convergent (but not necessarily alternating sign) series have this property, or is there one that diverges?

I feel like I'm missing some obvious triangle inequality trick, but I just don't see it.

$\endgroup$
0
8
$\begingroup$

No. Take your example, and interweave it with the all-zero sequence: $$ a_{n} = \begin{cases} \frac{(-1)^{n/2}}{n+1} &\text{ if } n \text{ even}\\ 0 &\text{ if } n \text{ odd} \end{cases} $$

$\endgroup$
8
  • 1
    $\begingroup$ Wow, I didn't think of that at all. $\endgroup$ Jan 14 '21 at 22:16
  • $\begingroup$ There might be more natural/less contrived, but it does the job. :) $\endgroup$
    – Clement C.
    Jan 14 '21 at 22:22
  • 2
    $\begingroup$ Note that this generalizes pretty broadly; the interwoven terms could e.g. be terms of an absolutely convergent series rather than zero, and they can be laid in whatever pattern is desired, to show that the answer is the same for whatever finite subsequence you might choose to average over. $\endgroup$ Jan 14 '21 at 22:28
  • $\begingroup$ @Steven I'm not sure I follow. $\endgroup$ Jan 14 '21 at 22:37
  • $\begingroup$ @AdamRubinson The counter example can be generalized to 3, or any fixed number of terms, instead of two consecutive; and some other patterns than just consecutive terms as well. $\endgroup$
    – Clement C.
    Jan 14 '21 at 22:39

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.