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Can an alternating series EVER be absolutely convergent?

I am examining practice problems in my calculus book and I haven't yet come across a case where this is so. It might be because they are simple, but I'm genuinely curious.

Don Shrinkle
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5 Answers5

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Hint: You could take any (absolutely) convergent series $\sum_{n=0}^\infty{a_n}$ where $a_n> 0$, and then consider $\sum_{n=0}^\infty{(-1)^na_n}$.

Hayden
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  • Ok, it would alternate between negative and positive numbers. Would it ever converge? – Don Shrinkle Apr 09 '16 at 05:34
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    For precision, you might want to say "take any absolutely convergent series". – marty cohen Apr 09 '16 at 05:35
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    @martycohen My intention was that because $a_n\geq 0$, convergence is the same as absolute convergence. I want to enforce $a_n\geq 0$ no matter what so that $(-1)^na_n$ can be guaranteed to be alternating, and guess I didn't want to write the additional word "absolutely" when it wasn't necessary. But yes, saying "absolutely convergent" makes the conclusion clearer. – Hayden Apr 09 '16 at 05:36
  • @DonShrinkle Do you know that absolute convergence implies convergence? – Hayden Apr 09 '16 at 05:36
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    If it had only non negative terms and it converges then it converges absolutely for obvious reasons. I really don't think it needs to be stated. In fact I think stating implies a stronger condition then the spirit of the argument requires. – fleablood Apr 09 '16 at 05:39
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    I think the only reason you haven't come across an alternating absolutely convergent series is that they aren't particularly interesting or illuminating. Here's an easy one. 1/2 - 1/4 + 1/8 -1/16... = 1/3. This is also equal to 1/4 + 1/16 + 1/64 +... So that it alternates and converges absolutely isn't particularly illuminating of anything. – fleablood Apr 09 '16 at 05:49
  • @bof, yes, that's a good point. I guess I didn't really think when I wrote $\geq$ instead of $>$. – Hayden Apr 09 '16 at 05:50
  • See, the thing is getting a series to converge is a big deal, knowing it converges absolutely is our huge gun that let's us breathe easy in the knowledge that whatever happens, we've got it covered. Nothing else matters; the terms can alternate signs or tap out a message in Morse code... we don't care-- we've assured convergence. So the only thing interesting about alternate series is that we don't need the huge gun. So the counterexamples of non absolute convergence are the interesting ones. – fleablood Apr 09 '16 at 05:59
  • Thinking about this yourself is a great exercise -- it is these sort of questions that are the rare and valuable ones. – Nate 8 Apr 09 '16 at 16:36
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$$ \sum_{n=0}^\infty \frac{(-1)^n}{2^n} = \frac 2 3. \qquad \sum_{n=0}^\infty \frac 1 {2^n} = 2. $$

Michael Hardy
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a series is absolutely convergent if $\sum |a_n| < M$

If a series is absolutely convergent then every sub-series is convergent.

Consider $\sum (-1)^n|a_n|$ The sum of the of the even terms converges, the sum of the odd terms converges.

user317176
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    I might have written "A series $\sum a_n$ is absolutely convergent if $\sum|a_n|<\infty$. Just what this quantity called $M$ is you don't say, and which series is absolutely convergent you don't say (you can't mean $\sum|a_n|$ is the one that's absolutely convergent). $\qquad$ – Michael Hardy Apr 09 '16 at 14:43
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Let $p_n > 0$ be a positive sequence and $a_n = (-1)^n p_n$ a corresponding alternating sequence.

Then in short, even if the series $P =\sum p_n$ does converge or not, the pair-wise cancelation power of having consecutive terms of opposite signs in the series allows $A=\sum a_n$ to converge more easily, under milder conditions on $p_n$.

For instance if $p_n \to 0$ monotonously, then $A$ converge, whatever $P$ does (converges or not). And you can easily bound the convergence rate. This is called the Leibniz rule.

For instance, the alternating harmonic series $\frac{(-1)^n}{n}$ has a finite sum ($-\log 2 \simeq 0.693$) but the harmonic series $\frac{1}{n}$ does not. A simple change of signs allows to turn an infinite (sum of the harmonic series) into a quite small number.

If $P$ converges (and absolutely by definition), then $A$ converges absolutely too and even more easily, as you can see since $|A| \le P$.

For instance, the geometric series $1/2 + 1/4 + 1/8 + 1/16 + ⋯$ sums to $1$ and its alternating counterpart $1/2 − 1/4 + 1/8 − 1/16 + ⋯$ sums to $1/3$.

Finally, alternating series are useful in practice, they can be used for faster numerical summation with series acceleration.

Laurent Duval
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If $\sum |a_n|$ converges, then $\sum \pm a_n$ converges for all choices of $+$ and $-$.

Steven Alexis Gregory
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