What is wrong with the following proof? This originates from the problem $\#3.8$ of Rudin, the problem is as followed:


  If $ \sum a_{n} $ converges and if ${b_{n}}$ is monotonic and bounded,
  prove $\sum a_{n}b_{n}$converges.


I know that if we remove the assumption that $b_{n}$ is monotonic, then the conclusion won't hold. That is, if we have $ \sum a_{n} $ converges and  ${b_{n}}$ bounded, we can not reach the conclusion that $\sum a_{n}b_{n}$converges. I think of a counter example as $a_{n}=\frac{(-1)^{n}}{n}$ and $b_{n}=(-1)^{n}$,then  $ \sum a_{n} $ converges and  ${b_{n}}$ bounded, however, $\sum a_{n}b_{n}$ diverges. 

Then my question is, however, I read of a proof  the problem without using the assumption that $b_{n}$ is monotonic, and hardly can I find where is wrong in the proof. So can anyone tell my what is wrong in the following proof? Thanks in advance guys!


  Exercise $\mathbf{3.8}$
  
  We're told that $\{b_n\}$ is bounded. Let $\alpha$ be the upper bound of $\{|b_n|\}$. We're also told that $\sum a_n$ converges: so for any arbitrarily small $\epsilon$, we can find an integer $N$ such that 
  $$\left|\,\sum_{k=m}^n a_k\right|\le\dfrac{\epsilon}{\alpha}\,\,\text{for all $n,m$ such that $n\ge m\ge N$}$$
  which is algebraically equivalent to 
  $$\left|\,\sum_{k=m}^n a_k\alpha\right|\le \epsilon\,\,\text{for all $n,m$ such that $n\ge m\ge N$}$$
  which since $|b_k|\le \alpha$ for every $k$, means that 
  $$\left|\,\sum_{k=m}^n a_k b_k\right|\le\left|\,\sum_{k=m}^n a_k \alpha\right|\le\epsilon\,\,\text{for all $n,m$ such that $n\ge m\ge N$}$$
  By theorem $3.22$, this is sufficient to prove that $\sum a_n b_n$ converges.
The last inequality is wrong. Just take $n=m+1$ and $a_n = -a_m$. Then $|\sum_{k=m}^na_k \alpha| = 0$, but it certainly doesn't follow that $|\sum_{k=m}^na_kb_k| = 0$ for all possible choices of $b_m, b_n$.
The first inequality is wrong.  It assumes that $\sum a_n$ converges absolutely, when the sum was only stated to converge.