A counterexample to monotonicity in Leibniz criterion for alternating 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:


*

*$\lim_{n\to+\infty}a_n =0$,

*$a_n$ 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 $a_n$.

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.
