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My attempt:

Since $(b_n)_{n \in \mathbb{N}}$ is bounded it follows that there exists a bounded sequence $$(c_n)_{n \in \mathbb{N}}: 0 \leq |b_n| \leq |c_n|, \forall n \in \mathbb{N}$$

Therefore it follows that: $$\sum_{n=1}^{\infty} a_nb_n \leq \sum_{n=1}^{\infty} a_nc_n$$

Since $\sum_{n=1}^{\infty}a_n$ converges absolutely $\Rightarrow$ $\sum_{n=1}^{\infty}a_n$ converges $\Rightarrow (a_n)_{n \in \mathbb{N}} $ is a null sequence $\Rightarrow \exists N \in \mathbb{N}: \forall n \geq N: |a_n| \lt \frac{\epsilon}{|c|p}$, where $c:= max\{c_1,\dots,c_n,\dots\}, p \in \mathbb{N}$

Then it follows that: $$\forall n \geq N: |a_nc_n|=|a_n||c_n| \lt |a_n||c| \lt \frac{\epsilon}{|c|p}|c|=\frac{\epsilon}{p}$$

$$\Rightarrow \forall n \geq N: |a_{n+1}c_{n+1}+\dots+a_{n+p}c_{n+p}|\leq |a_{n+1}c_{n+1}|+ \dots +|a_{n+p}c_{n+p}| \lt p\frac{\epsilon}{p}=\epsilon$$

So after Cauchy's convergence test, it follows that $\sum_{n=1}^{\infty} a_nc_n$ is a convergent majorant of $\sum_{n=1}^{\infty} a_nb_n$ and therefore $\sum_{n=1}^{\infty} a_nb_n$ converges absolutely.

Was this proof correct?

And a second part of that question is: If $\sum_{n=1}^{\infty}$ converges, but not absolutely, does $\sum_{n=1}^{\infty} a_nb_n$ converge?

For my proof I didn't really needed $\sum_{n=1}^{\infty}$ to converge absolutely, I just used the fact that it converges when it converges absolutely and since I proved that $\sum_{n=1}^{\infty} a_nb_n$ converges absolutely, it converges as well. Is that right?

eager2learn
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    You are over thinking it. The sequence $b_n$ bounded means there is an $M$ so that $|b_n| \leq M$ for all $n$. Then $\leq |a_n b_n| \leq M |a_n|$ and we conclude convergence by the comparison test since $a_n$ converges absolutely. Done – abnry May 31 '14 at 14:36
  • I have one question though, I don't need absolute convergence for my proof, it's enough to assume that $(a_n)$ converges to show that $\sum_{n=1}^{\infty}a_nb_n$ converges absolutely. So I assume I have made a mistake in my proof. Can anybody point out where the mistake is? – eager2learn Jun 01 '14 at 09:51

1 Answers1

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If $\{b_n\}$ is a convergent sequence, then there exists an $M$ such that $|b_n|\le M$ for all $n$. It then follows that $$ \sum_{n=0}^{\infty}|a_nb_n| = \sum_{n=0}^{\infty}|a_n||b_n|\le \sum_{n=0}^{\infty}|a_n|M \le M\sum_{n=0}^{\infty}|a_n| < \infty. $$ The second part is not true. Take $a_n = (-1)^n\frac{1}{n}$ and $b_n = (-1)^n$, then $\{b_n\}$ is bounded, but $$ \sum_{n=1}^{\infty}a_nb_n = \sum_{n=1}^{\infty}\frac{1}{n}, $$ which diverges.

Marc
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