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I've seen the example of a divergent alternating series given as follows:

$$\sum_{n=1}^{\infty} \frac{1}{n^2} - \frac{1}{n}$$

This fails the test for absolute convergence, so testing for conditional convergence:

  1. $$\lim_{n \to \infty} \frac{1}{n^2} - \frac{1}{n} = 0$$

That passes.

However, taking the derivative of the absolute value of the term function:

  1. $$ f(x) = \lvert \frac{1}{x^2} - \frac{1}{x}\rvert $$

But $f'(x) < 0 $ for $x > 2$, so this is monotonically decreasing. Meeting these two conditions means it passes the alternating series test for absolute convergence. But where did I go wrong, as this certainly diverges?

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    $\begingroup$ "alternating series" In what way does this alternate? Every term seems to be negative to me. $\endgroup$ – JMoravitz Apr 7 '17 at 5:57
  • $\begingroup$ Please see this link $\endgroup$ – rb612 Apr 7 '17 at 5:58
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    $\begingroup$ So, you have interlaced the sequence $\frac{1}{n^2}$ with the sequence $-\frac{1}{n}$. The terms then alternate between the two sequences. If you want to apply any sort of test, you should apply it to one term followed by the next term, not to pairs of terms simultaneously followed by the next pair of terms. $\endgroup$ – JMoravitz Apr 7 '17 at 6:00
  • $\begingroup$ Ah, so is that the reason for me getting an inconsistent result, that it conditionally converges when it really doesn't? What is a good test to apply to a problem like this? $\endgroup$ – rb612 Apr 7 '17 at 6:03
  • $\begingroup$ Here, your $f(n)$ is not $\frac{1}{n^2}-\frac{1}{n}$ if you want to call this an alternating series, but rather something more along the lines of $\frac{1}{2n^2}+(-1)^n\frac{1}{2n^2} - \frac{1}{2n-2}+(-1)^{n+1}\frac{1}{2n-2}$, making some corrections to the specific function to account for the "n repeating once before moving on to the next." Its not interesting enough of a question for me to fix the function to work exactly. The point is, if your $f(n)$ is literally $\frac{1}{n^2}-\frac{1}{n}$, then every term is negative and it does not in fact alternate. $\endgroup$ – JMoravitz Apr 7 '17 at 6:04
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If we denote $\sum_{n=1}^M \frac {1}{n^2}$ by $a_M$ and $\sum_{n=1}^M \frac {1}{n}$ by $b_M$ then we know that $a_M$ is a sequence converging towards a real number (which is $\frac {\pi^2}{6}$ but that's not relevant) and also that $b_M$ is a diverging sequence going towards $+\infty$. Combining these results it is clear that the sequence $a_M-b_M$ diverges towards $-\infty$. And because $$ a_M - b_M = \sum_{n=1}^M \frac{1}{n^2} - \frac{1}{n},$$ we have $$\sum_{n=1}^{+\infty} \frac{1}{n^2} - \frac{1}{n} \to -\infty.$$

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  • $\begingroup$ Cool, thanks! so we can't apply the Alternating Series Test here because the combined expression doesn't alternate above and below zero? $\endgroup$ – rb612 Apr 7 '17 at 12:53
  • $\begingroup$ That's right all terms but the first are negative so it's not alterning. $\endgroup$ – user322559 Apr 7 '17 at 14:03

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