Cfrgtkky

I mean if $a_k>b_k$ for every k and we know the sum of b_k diverges, ie the sequence of partial sums of b_k go to infinity, then since each a_k>b_k, the partial sums of a_k are always bigger than those of b_k, hence the partial sums of a_k should go to infinity as well? Thanks.

edit: by sum of b_k diverges I mean assuming the sum of b_k is infinity. i dont see how sum a_k can be anything other than infinity in this case as well.

Easy to see that $sum 0$ converges, and
$$
0 = sum 0 geqslant sum -n^{-1}, 0 geqslant sum -n^{-2},
$$
but $sum -n^{-1} = -infty$, $sum -n^{-2} = - sum n^{-2}$ converges.

Comparison test works because the sequence of partial sums are monotonically increasing, so when $0leqslant a_n leqslant b_n$, the partial sums $A_n leqslant B_n$, then when $sum b_n$ converges, $B_n < +infty$ implies $A_n < +infty$, and $A_n$ converges because it is bounded above by $sum b_n$.

UPDATE

Divergence does not only mean that the partial sum goes to infinity, a typical counterexample is $1-1+1-1+1-1+cdots$: the partial sums are bounded by $1$. If you know that $B_n to +infty$, then you could conclude that $A_n to +infty$ given that $a_n geqslant b_n$. Otherwise you should not conclude anything before you investigate $b_n, a_n$ further.

Consider the sequences

$a_k=frac{{(-1)}^{k+1}}{k}$ $, kin mathbb{N}$

$b_k= begin{cases} -1,& text{if k is even } \ 0, & text{if k is odd } end{cases}$

Here $a_k > b_k$ $forall kin mathbb{N}$

Also, $sum_{k=1}^infty b_n$ is not convergent here but $ sum_{k=1}^infty a_k$ is convergent

This shows why comparison test not applicable when terms are non-negative.
However, it is applicable in case if all but finitely many terms are non-negative.