# A counterexample to monotonicity in Leibniz criterion for alternatig series

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 $a_n$.

• Do you have any ideas? – Mark Viola Jun 10 '18 at 15:19
• No, I haven't... – zar Jun 10 '18 at 15:29

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.