This is a follow-up from this question.

Hypothesis: If $\sum\limits_{n=0}^{\infty} a_n\ $ is a conditionally convergent series with $\ \limsup_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = 1$, then is $\sum\limits_{n=0}^{\infty} |a_n+a_{n+1}|\ $ convergent?

Maybe I'm wrong but I believe this isn't answered in the comments to the answer, because not only does it not allow for zeros to be part of the sequence, but also it's not clear to me how interweaving an absolutely convergent sequence answers this question. $$$$ Actually I think I've found a sequence that disproves my hypothesis. For the following sequence, it doesn't matter what $a_0, a_1$ are: $a_0 =-1,\ a_1 =-1,\ a_2 = \frac{1}{2},\ a_3 = \frac{1}{3},\ a_4 = -\frac{1}{4},\ a_5 = -\frac{1}{5},\ a_6 = \frac{1}{6},\ a_7 = \frac{1}{7},\ a_8 = -\frac{1}{8},\ a_9 = -\frac{1}{9}, ...$

It's an alternating sequence of decreasing terms (which $\to 0$ ) when put into pairs $(a_0 + a_1, a_2+a_3, a_4+a_5,... ),$ so I think the corresponding series $\sum\limits_{n=0}^{\infty} a_n\ $ is conditionally convergent. Also, the series $\sum\limits_{n=0}^{\infty} |a_n+a_{n+1}|\ $ is $\geq $ the sum of the pairs: $\sum\limits_{ \text{n is even} }^{\infty} |a_n+a_{n+1}|\ \geq \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + ...,$ which famously diverges. So I believe this sequence disproves my hypothesis(but I'd like confirmation please). $$$$ Assuming this is all correct, this then begs the modification to the original question:

If $\sum\limits_{n=0}^{\infty} a_n\ $ is a conditionally convergent series with $\ \limsup_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = 1$, then is $\sum\limits_{n=0}^{\infty} a_n+a_{n+1}\ $ convergent?

But for this question, we can even get rid of the $\ \limsup_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = 1$ condition and the answer is still yes, because

$\sum\limits_{n=0}^{\infty} a_n+a_{n+1}\ = lim_{n \to \infty} \left(a_0+2\left(a_1+a_2+a_3+...a_{n-1}\right) + a_n\ \right) = 2\sum\limits_{n=0}^{\infty} a_n\ - a_0, $ which converges because $\sum\limits_{n=0}^{\infty} a_n\ $ converges because it is conditionally convergent.


1 Answer 1


Conditional convergence means the sum converges, but not absolutely.

That is, $\sum_{n\in \Bbb N} a_n $ converges, but $\sum_{n\in\Bbb N} |a_n|$ does not.

Now: $\limsup_{n\to\infty} |\frac{a_{n+1}}{a_n}|=1$ tells us that $|a_{n+1}|$ and $|a_n|$ are very close in value as $n\to\infty$, which implies we can take: $$\sum_{n\in\Bbb N} |a_n+a_{n+1}|\approx\sum_{n\in\Bbb N} |2a_n|=2\sum_{n\in\Bbb N} |a_n|$$ which does not converge.

For the same reason: $$\sum_{n\in\Bbb N} a_n+a_{n+1}\approx \sum_{n\in\Bbb N} 2a_n=2\sum_{n\in\Bbb N} a_n$$ whenever $a_n$ and $a_{n+1}$ are sufficiently close, this implies convergence. While $\limsup|\frac{a_{n+1}}{a_n}|=1$ is a sufficient condition for this, it is not necessary.

  • $\begingroup$ If you're saying that $\limsup_{n\to\infty} |\frac{a_{n+1}}{a_n}|=1$, implies $\sum_{n\in\Bbb N} |a_n+a_{n+1}|\ $ diverges, then you're wrong. For example, the alternating harmonic series has $\limsup_{n\to\infty} |\frac{a_{n+1}}{a_n}|=1$, and $\sum_{n\in\Bbb N} |a_n+a_{n+1}|\ $ is half the reciprocal of the triangular numbers, which converges. $\endgroup$ Jan 16, 2021 at 12:47

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