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.
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$\begingroup$Hint: You could take any (absolutely) convergent series $\sum_{n=0}^\infty{a_n}$ where $a_n> 0$, and then consider $\sum_{n=0}^\infty{(-1)^na_n}$.
$\endgroup$ 9 $\begingroup$$$ \sum_{n=0}^\infty \frac{(-1)^n}{2^n} = \frac 2 3. \qquad \sum_{n=0}^\infty \frac 1 {2^n} = 2. $$
$\endgroup$ $\begingroup$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.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ $\begingroup$If $\sum |a_n|$ converges, then $\sum \pm a_n$ converges for all choices of $+$ and $-$.
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