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I'm trying to convince myself that all three conditions

\begin{equation} \tag{1} (a_n)_{n\in\mathbb{N}} \text{ is null sequence} \end{equation} \begin{equation} \tag{2} (|a_n|)_{n\in\mathbb{N}} \text{ is monotonically decreasing} \end{equation} \begin{equation} \tag{3} a_n \text{ has alternating signs} \end{equation}

of the Leibniz criterion are needed by finding examples where one condition is violated and $\sum a_n$ doesn't exist.

For (1), I've got $a_n = (-1)^n$ (meets (2) and (3) but $\sum (-1)^n$ doesn't exist).

For (3), I've got $a_n = \dfrac{1}{n}$ (meets (1) and (2) but $\sum \dfrac{1}{n}$ doesn't exist).

But I can't find a sequence that meets (1) and (3) but not (2) so that $\sum a_n$ diverges. I'm pretty sure that it needs to contain a $(-x)^n$ part of some sort to meet (3) but as soon as I make it small enough to be a null sequence (e.g. by dividing by $n$) the series converges.

Clayton Louden
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1 Answers1

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Hint: interleave a convergent series of negative terms with a divergent series of positive terms.

Rob Arthan
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  • Could you give an additional hint on interleave? Could the alternating sign come from adding something and then substracting something again? – Clayton Louden Dec 12 '15 at 03:02
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    Interleave = ripple shuffle = dovetail = make alternate choices from the positive terms and the negative terms (no additions or subtractions required). – Rob Arthan Dec 12 '15 at 03:13
  • That's so simple! I didn't even consider that. It's almost like cheating! :) Thank you very much – Clayton Louden Dec 12 '15 at 03:48