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Reason I ask I know a series can converge but then when you apply the absolute convergence test it may diverge. I understand this part. One concludes absolute convergence is a stronger condition!

But what happens if the original series diverges and the terms are negative , how do I know that by making it positive it won't become convergent? In this scenario you would never know that absolute convergence was the strongest condition.

The solution would be that it makes no mathematical sense to apply absolute convergence to a divergent series. Or is this just by definition maybe?

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    $\begingroup$ Sorry, maybe I didn't understand your doubt but isn't $\sum a_n$ diverges, with $a_n$ negative $\implies \sum |a_n|=-\sum a_n$ also diverges? $\endgroup$ – gimusi Feb 12 '18 at 17:34
  • $\begingroup$ The contrapositive of the statement “any absolutely convergent series is also convergent” is “any divergent series is also absolutely divergent”, which is of course true. It perfectly makes sense to apply this statement, but simply it adds no new information on your divergent series. $\endgroup$ – Sangchul Lee Feb 12 '18 at 17:38
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As you know and stated,

If a series is absolutely convergent, then it is convergent.

By contrapositive:

If a series is divergent, then it is not absolutely convergent.

So indeed, performing an absolute convergence test on a divergent series is pointless: you know the answer already.


For your specific point (to which the above of course applies, but can be dealt with more specifically as well): theorems for positive series apply to negative series as well. This is simply because $\sum_n a_n$ converges iff $\sum_n (-a_n)$ converges.

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  • $\begingroup$ is that not exactly my point? $\endgroup$ – Guy Fsone Feb 12 '18 at 17:40
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    $\begingroup$ @GuyFsone The point you made while I was typing my answer and that I only saw after I posted it (and doesn't include the last paragraph)? Yes, I reckon. $\endgroup$ – Clement C. Feb 12 '18 at 17:42
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If the original series diverges then it cannot possibly be absolutely convergent.

Hint: Cauchy criterion for convergence. For all $m > n$ sufficiently large

$$\left|\sum_{k=n}^m a_k\right| \leqslant \sum_{k=n}^m |a_k| < \epsilon$$

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  • $\begingroup$ is your argument similar to mine, or I am wrong? Thanks $\endgroup$ – gimusi Feb 12 '18 at 17:44
  • $\begingroup$ @gimusi: Mine is just a hint. There is nothing wrong with yours or any of these answers. I added an upvote to your answer. $\endgroup$ – RRL Feb 12 '18 at 18:14
  • $\begingroup$ Thanks I didn’t understand the downvotes and I was wondering if I was missing something in the OP. Thanks again! $\endgroup$ – gimusi Feb 12 '18 at 18:17
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Note that the following mathematical propositions are equivalent

$$(P\implies Q)\equiv (\neg Q\implies \neg P)$$

Now take

$P : $ (The series $\sum(a_n)_n$ converges absolutely )

and

$Q: $ ( The series $\sum(a_n)_n$ converges )

and we have,

((The series $\sum(a_n)_n$ converges absolutely ) $\implies$ The series $\sum(a_n)_n$ converges) )

Hence we have

((The series $\sum(a_n)_n$ diverges ) $\implies$ The series $\sum(a_n)_n$ diverges absolutely ) )

what can you conclude?

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Note that if $\sum a_n$ diverges, with $a_n$ negative, also $\sum |a_n|=-\sum a_n$ diverges.

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  • $\begingroup$ Could you please also explain what is wrong in this simple observation? Thanks $\endgroup$ – gimusi Feb 12 '18 at 17:48
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Let $a_n$ be any divergent series. Define $$b_n=\left\{ \begin{array}{lc} a_n & \mbox{ if } a_n \geq 0 \\ 0& \mbox{ if } a_n < 0 \\ \end{array} \right. \\ c_n=\left\{ \begin{array}{lc} -a_n & \mbox{ if } a_n < 0 \\ 0& \mbox{ if } a_n \geq \\ \end{array} \right. \\$$

Then, $\sum b_n, \sum c_n$ are non-negative series and $$\sum_{k=1}^n a_k =\left(\sum_{k=1}^n b_k \right)-\left(\sum_{k=1}^n c_k \right) \\ \sum_{k=1}^n |a_k| =\left(\sum_{k=1}^n b_k \right)+\left(\sum_{k=1}^n c_k \right) $$

Since $b_n, c_n$ are non-negative, the following results are obvious:

Lemma 1 If $\sum_{k=1}^\infty |a_k| <\infty$ then $$\sum_{k=1}^\infty b_k <\infty \\ \sum_{k=1}^\infty c_k <\infty$$

Lemma 2: If $\sum_{k=1}^\infty b_k, \sum_{k=1}^\infty c_k$ are convergent, then so is $$\sum_{k=1}^n a_k =\left(\sum_{k=1}^n b_k \right)-\left(\sum_{k=1}^n c_k \right) $$

It is easy to conclude from here that divergence of the series implies that $\sum_{k=1}^\infty |a_k|=\infty$.

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