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Leibniz criterion for alternating series says that the series $$ \sum_{n=0}^{+\infty}(-1)^n a_n $$ conditionally converges if the following two hypothesis are verified:

  1. $\lim_{n\to+\infty}a_n =0$,
  2. $a_n$ is monotonically decreasing.

My question is: if $b_n$ is asymptotic to $a_n$ (which means that $\lim_{n\to+\infty}a_n/b_n=1$), can I study the monotonicity of $b_n$ instead of $a_n$?

I know the proof of the Leibniz criterion, and I am aware of the fact that if $b_n$ is asymptotic to $a_n$ this doesn't mean that $b_n$ is monotonically decreasing.

What I'm searching for is a counterexample: I would like to find a non converging alternating series $\sum_{n=0}^{+\infty}(-1)^na_n$ and a (conditionally) convergent alternating series $\sum_{n=0}^{+\infty}(-1)^nb_n$ where $b_n$ is asymptotic to $a_n$.

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  • $\begingroup$ Do you have any ideas? $\endgroup$
    – Mark Viola
    Jun 10, 2018 at 15:19
  • $\begingroup$ No, I haven't... $\endgroup$
    – zar
    Jun 10, 2018 at 15:29

1 Answer 1

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Consider the sequences $$a_n:={1\over n}+{(-1)^n\over n\>\log n}, \quad b_n:={1\over n}\qquad (n\geq2)\ .$$ Then ${a_n\over b_n}\to1$, the series $\sum_{n=2}^\infty(-1)^n b_n$ is convergent, but the series $\sum_{n=2}^\infty (-1)^n a_n$ is not.

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  • $\begingroup$ Oh, thank you, now I see the problem. $\endgroup$
    – zar
    Jun 10, 2018 at 15:30

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