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I have read the correct proof, which makes sense. But, I want to know if the following proof is wrong:

If $\sum a_n$ converges, and if $\{b_n\}$ is monotonic and bounded, prove that $\sum a_n b_n$ converges.

Proof. Since $\{b_n\}$ is bounded, by the lowest upper bound property, it has a supremum, say $b$. Since $\sum a_n$ converges, $\forall\varepsilon >0, \exists N$ s.t. $$\left|\sum_{k=n}^ma_k\right| \le \frac{\varepsilon}{c},$$ where $c=\max\{|b|,1\}$ and $m\ge n\ge N$. Then, $$\left|\sum_{k=n}^m a_k b_k\right|\le\left|b\sum_{k=n}^ma_k\right| = |b|\cdot\left|\sum_{k=n}^m a_k\right|\le\varepsilon.$$ Hence, $\sum a_nb_n$ converges.

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To find the error yourself: note that you never used the assumption that $b_n$ was monotonic. So if your proof were correct, it would yield the conclusion when $b_n$ is only bounded. But the theorem is clearly not true in that case; take $a_n$ to be the alternating harmonic series and $b_n = (-1)^n$. Now see which line of your proof is false for this example.

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Second last line: Why does the left inequality hold?

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  • $\begingroup$ That's indeed the problem. Thank you. $\endgroup$ – Xward Mar 4 '18 at 6:28

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