The comparison test for series tells us (quoted from wikipedia):

"If the infinite series $\sum b_n$ converges and $0 \leq a_n \leq b_n$ for all sufficiently large $n$ (that is, for all $n>N$ for some fixed value $N$), then the infinite series $\sum a_n$ also converges.

If the infinite series $\sum b_n$ diverges and $0 \leq b_n \leq a_n$ for all sufficiently large $n$, then the infinite series $\sum a_n$ also diverges."

But it doesn't tells us anything abou what happens when $b_n$ is a nonnegative sequence and $\sum b_n$ diverges and another nonnegative sequence $a_n$ is crossing up and down infinitely many times $b_n$, I mean when there are infinite $n$'s where $a_n < b_n$ and also infinite $n$'s where $a_n \geq b_n$. Intuitively it seems to me that in this case the series $a_n$ must diverge since I can split the sequence $a_n$ into two subsequences, one having all the terms $a_n \geq b_n$ (lets call it $c_n$) and the other one having all the terms $a_n<b_n$ (lets call it $d_n$, so the series $\sum a_n = \sum c_n + \sum d_n$. Now because of the second statement of the comparison test the series $\sum c_n$ diverges and so does $\sum a_n$.

Is this reasoning right? Thanks in advance

• Your argument does not work: why do you conclude that $\sum_n c_n$ is divergent? See my counterexample in the answer below. Apr 19 '16 at 12:33
• @Crostul deleted! Thanks for the clarification! Apr 19 '16 at 15:19

Your argument does not work for non-monotone sequences. For example, consider $$b_n = \begin{cases} \frac{1}{n} & \mbox{if n is even} \\ \frac{1}{n!} & \mbox{if n is odd} \end{cases}$$ and $a_n = \frac{2}{n!}$.
Clearly $\sum_n b_n$ diverges (consider the subsequence given by even indices), while $\sum_n a_n$ converges. Now: for all $n$,
1. If $n$ is even, $a_n < b_n$ :but this does not give us any information about divergence of $\sum_{n \mbox{ even}} a_n$
2. If $n$ is odd, $a_n > b_n$ : but this does not give us any information about divergence of $\sum_{n \mbox{ odd}} a_n$