First of all argument bounded and monotone will not work because there is no restriction on sequences to be positive or negative, they can alternate irregularly.
I tried to make a use of comparison test but there was too many cases to consider and it was just too messy.
So I think the only way to approach the problem is directly from the definition. All I need to do is to prove that the sequence of partial sums $B_n=\sum^n_{i=0} b_i$ is convergent sequence. So I tried to prove that $B_n$ is Cauchy because this seemed easier. $$a_n<b_n<c_n \\ \therefore \ a_n+a_{n+1}+...+a_{n+m}<b_n+b_{n+1}+...+b_{n+m}<c_n+c_{n+1}+...+c_{n+m}\\ \therefore \ A_{n+m}-A_n<B_{n+m}-B_n<C_{n+m}-C_n$$ and by sandwich this goes to zero $B_{n+m}-B_n$ goes to zero as $n$ tend to infinity. The problem is that I need no this expressions but modulus of them and in this case it does not need to be true for example if $A_{n+m}-A_n<0<B_{n+m}-B_n$. And here I am stuck
The partial sums form a Cauchy sequence: If $n<m$ then $$\sum_{j=n}^m a_j < \sum_{j=n}^m b_j<\sum_{j=n}^m c_j.$$Since $\sum_{j=n}^m a_j\to0$ and $\sum_{j=n}^m c_j\to0$ as $n,m\to\infty$ it follows that $\sum_{j=n}^m b_j\to0$, qed.