Existence of additional condition for Convergence of $\sum a_nb_n$ with monotonicity condition is dropped?

Question

I know there are 2 theorem for convergence of $\sum a_nb_n$
which has following assumption.
1) If $\sum a_n$ is convergent and $b_n$ is monotonic and bounded. then $\sum a_nb_n$ is convergent.
2) If $\sum a_n$ is bounded and $b_n$ is monotonic and convergent to 0. then $\sum a_nb_n$ is convergent.
I know that we can not drop monotonicity condition on both as there are counter example are occuring.
I am thinking is there exist additional strong condition on $\sum a_n$ so that we can dropped monotonicity condition of $b_n$ I thought for absolute convergence of $\sum a_n$.
But for writting proof I could not get bound.
Is there exist some theorem in this regeard.
Any Help will be appreciated

Answer 1

Well, if $a_n$ are all positive, and $\sum a_n$ is convergent, and $b_n$ is bounded, then you have $$\sum a_n b_n \leq \sum a_n\cdot M \leq M\cdot \sum a_n$$ which means $\sum a_n b_n$ is convergent.

It's also enough to demand that $a_n$ is absolutely convergent.

Answer 2

Yes, that would be correct. That is, if the series $\sum_{n=0}^\infty a_n$ converges absolutely and if $\lim_{n\to\infty}b_n=0$, then $\sum_{n=0}^\infty a_nb_n$ converges. Actually, it converges absolutely. Since $\lim_{n\to\infty}b_n=0$, the sequence $(b_n)_{n\in\mathbb N}$ is bounded. Let $B$ be an upper bound of $\bigl(\lvert b_n\rvert\bigr)_{n\in\mathbb N}$. Then $(\forall n\in\mathbb{Z}^+):|a_nb_n|\leqslant B|a_n|$ and therefore $\sum_{n=0}^\infty\lvert a_bb_n\rvert$ converges, by the comparison test.

Answer 3

If $\sum_{n\ge1} a_n$ is absolutely convergent, i.e. $$ \|a\|_{\ell^1} := \sum_{n\ge1} |a_n| < \infty$$ and $b_n$ is bounded, $$ \|b\|_{\ell^\infty} := \sup_{n\ge1} |b_n| < \infty$$ then the sum $\sum_{n\ge1} a_n b_n$ absolutely converges, and we have the explicit bound $$ \sum_{n\ge1} |a_n b_n| \le \|a\|_{\ell^1}\|b\|_{\ell^\infty} < \infty$$

Actually, this is a special case of Holder's inequality (MSE post with proof, Wikipedia statement) on $\ell^p$, the spaces of "$p$-summable sequences": define additionally for all $p\in(1,\infty)$, $$\|a\|_{\ell^p} := \sqrt[p]{\sum_{n\ge1} |a_n|^p}$$ And define the following spaces for all $p\in[1,\infty]$, $$\ell^p := \{ \text{sequences }x = (x_n)_{n\ge 1} : \|x\|_{\ell^p} < \infty \}$$ Then if $p\in[1,\infty]$, $a\in\ell^p$ and $b\in \ell^q$, where $q$ is the "Holder conjugate", i.e. the number $q\in[1,\infty]$ solving$^*$ $$ \frac1p + \frac1q = 1$$ Then $\sum_{n\ge 1} a_n b_n$ is absolutely convergent, with the explicit bound $$ \sum_{n\ge1} |a_n b_n| \le \|a\|_{\ell^p}\|b\|_{\ell^q}$$

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${}^{*\text{ when you are only dealing with numbers $\ge 0$ like $p,q$ here, the slightly abusive $\frac{1}{\infty} = 0$ is useful.}}$