Show that if $\displaystyle \sum$ a_k is absolutely convergent and abs(b_k) $\displaystyle \leq$ abs(a_k) fopr all k, then $\displaystyle \sum$ b_k is absolutely convergent

Here is what I have for this:

Because abs(b_k) $\displaystyle \leq$ abs(a_k), then $\displaystyle \sum$ abs(b_k) $\displaystyle \leq$ $\displaystyle \sum$ abs(a_k)

With the basic comparison test, since $\displaystyle \sum$ abs(a_k) is convergent, then $\displaystyle \sum$ abs(b_k) is convergent

Since $\displaystyle \sum$ abs(b_k) is convergent, by theorem $\displaystyle \sum$ bk converges and is absolutely convergent.

Is this proof sufficient?