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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.

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The last inequality is wrong. Just use your example. – Arctic Char Dec 6 '13 at 9:06
But the OP's counterexample doesn't fit the hypotheses, because $b_n$ is not monotonic. – TonyK Dec 6 '13 at 9:07
Where does the proof use the fact that $b_n$ is monotone? – Lost Dec 6 '13 at 9:10
@Lost Thanks Lost, my question is that the proof doesnt use the hypothese that $b_{n}$ is monotonic and it seems to give a specious proof, and at first I did not find where is wrong. – Brain Zhang Dec 6 '13 at 9:15
up vote 4 down vote accepted

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 $b_m, b_n$.

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A bit of correction there - I don't believe this is Rudin's work, but rather someone else's proof that the OP was reading. – Lost Dec 6 '13 at 9:11
The proof isnt Rudin's work, its someone else's work! – Brain Zhang Dec 6 '13 at 9:13
OK, thanks. I've edited my answer now. – TonyK Dec 6 '13 at 9:14
@TonyK Thanks, Tony! very helpful! – Brain Zhang Dec 6 '13 at 9:18

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