# Can an alternating series ever be absolutely convergent?

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.

• Sure. $0+0+0+0+0+0+\dots$ Apr 9, 2016 at 17:06
• @PyRulez some definitions suppose non-zero terms Apr 9, 2016 at 20:32

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

• Ok, it would alternate between negative and positive numbers. Would it ever converge? Apr 9, 2016 at 5:34
• For precision, you might want to say "take any absolutely convergent series". Apr 9, 2016 at 5:35
• @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. Apr 9, 2016 at 5:36
• 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. Apr 9, 2016 at 5:39
• 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. Apr 9, 2016 at 5:49

$$\sum_{n=0}^\infty \frac{(-1)^n}{2^n} = \frac 2 3. \qquad \sum_{n=0}^\infty \frac 1 {2^n} = 2.$$

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.

• 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). $\qquad$ Apr 9, 2016 at 14:43

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.

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