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I need a hint for a practice problem:

Let $a_n \geq 0$. Show that if $\displaystyle\sum_{n=1}^\infty a_nb_n$ converges for every monotonically decreasing sequence $b_n \to 0$, then $\displaystyle\sum_{n=1}^\infty a_n$ converges.

I've been trying to use the fact that $\sum\limits_{n=1}^\infty a_n r^n \leq M < \infty$ for all $r \in [0,1)$ iff $\displaystyle\sum_{n=1}^\infty a_n$ converges, but I can't seem to get it, so I'm not sure that's the right way to go about it.

Thanks in advance!

Jonas Meyer
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user93370
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    Sorry, fixed it I think. $b_n \geq 0$ and decreasing monotonically to $0$. – user93370 Sep 06 '13 at 02:47
  • The first thing to notice is that your condition does imply that a$_n$ goes to zero. To see this consider that $\sum 1/n$ – Betty Mock Sep 06 '13 at 03:37
  • continued $\sum 1/n$ does not converge, so if $\sum a_n b_n$ converges you must have $a_n < 1/n$. That shows $a_n$ goes to zero, which is a necessary condition for convergence. An no matter fast $\sum b_n$ diverges, $a_n$ is small enough to force convergence. That is not a proof yet, but a pointer in a good direction. – Betty Mock Sep 06 '13 at 03:48
  • @BettyMock: I have edited your comment so it renders properly (as you might have expected, there was a missing $). – user642796 Sep 06 '13 at 04:26
  • I expected there was a wrong something. I appreciate you cleaning this up. I would have fixed it myself, but couldn't get to it. – Betty Mock Sep 07 '13 at 03:30
  • This is related to the fact that $\ell^1$ is the dual space of $c_0$. – Jonas Meyer Jul 27 '14 at 16:55

2 Answers2

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(Contraposition.) Suppose that $a_n\geq 0$ for each $n$ and $\sum\limits_{n=1}^\infty a_n=+\infty$. Define $n_1<n_2<n_3<\cdots$ such that $\sum\limits_{n=n_{k}}^{n_{k+1}-1}a_n>1$ for each $k$ (with $n_0=1$). Define $b_n=\frac{1}{k}$ for $n_{k-1}\leq n<n_k$. Then $\sum\limits_{n=1}^\infty a_nb_n$ diverges.

Jonas Meyer
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$\sum a_nb_n$ converges for every $b_n$ which goes monotonically to 0. Let $b_n$ be such a sequence such that $\sum b_n$ diverges. If $ \sum a_n b_n$ is to converge, then "on average" the {$a_n$} < the {$b_n$}, even though for any particular n it could be that $a_n > b_n$. One way to write this is

$ \sum_{1}^n a_n$ < $\sum_{1}^n b_n$ for "most" n. More precisely as $n \rightarrow \infty$ we have $\sum a_n$ < $ \sum b_n$. That is, $\sum a_n$ is strictly less than $\sum b_n$ whenever b is divergent.

We can define a series {$b_n$} as being divergent in this way: that for any number R > 0 there exists a number $N_R$ such that $ \sum_{n =1}^{N_R} b_n> R$.

Let {$b_n$} be a divergent series and pick R. The $N_R$ that works for {$b_n$} will not work for {$a_n$}. The reason is that for every positive integer k, $\sum_{n=1}^\infty a_n$ < $ \frac{1}{k} \sum _{n=1}^\infty b_n$ ,(which is divergent) and there is always a k such that the chosen $N_R$ is too small, unless $\sum a_n$ converges to a number > R.

Since $ \frac{1}{k} \sum _{n+1}^\infty b_n$ does diverge, there is a new $N_{R,k}$ which works for this series, and if we take k large enough we have $N_{R,k}$ > $N_R$ where the inequality is strict.

However, no matter what k we pick $N_{R,k}$ if $\sum a_n$ does not converge to a number > R it is not large enough that $\sum_{n = 1}^{N_{R,k}} a_n> R$, because there is always a larger k, thus a smaller divergent series, and thus a bigger $N_{R,k}$ for that series.

If for any R there can be no ${N_R}$ for $\sum a_n$ then that sum cannot be divergent. If it is not divergent, it must converge.

Betty Mock
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  • In the future, if you find that an answer of yours is not quite up to snuff, instead of replacing the text with something that is surely not an answer, simply delete the answer, and if you later find a way to fix the answer you can then edit and undelete the answer yourself. Cheers! – user642796 Sep 08 '13 at 05:14
  • @ArthurFischer -- I'm kind of new to the site, and haven't figured everything out yet. After I saw that your delete locked it, I realized I should have done what you just said. Thanks for your patience. – Betty Mock Sep 09 '13 at 17:04
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    In the first paragraph: "we must have $a_nn^q<1$"-- you cannot say that holds for all $n$, but at least for infinitely many; the set of $n$ where that inequality does not hold may also be infinite. In the second paragraph: "so that $a_n<1/n$ and $\lim a_n<1/n$ as $n\to\infty$ for all n..." I cannot make sense of that. There is a limit on one side of the inequality, an expression with $n$ on the other side, and "$n\to\infty$ for all $n$..." must not be intended. In any case, $a_n<1/n$ for all $n$ does not imply convergence, e.g. $a_n=1/(2n)$. – Jonas Meyer Sep 17 '13 at 11:38
  • @JonasMeyer the expression $a_nn^q$ cannot have infinitely many terms greater than 1 or the series $\sum 1/n^q$ where q > 1 will not converge; and we know it does. This statement holds for every q >1. So in the limit as $q \rightarrow 1 a_n \le 1/n$. It is true that $a_n = 1/(2n)$ gives $a_n < 1/n$. However, this isn't about a constant factor. It's about forcing $a_n < 1/(n^q)$ for some q >1, or if you wish $ a_n < k/(n^q)$ where k is any positive integer. That has to be true or we can construct a $b_n$ going monotonically to 0 such that $\sum a_nb_n$ does not converge. – Betty Mock Sep 17 '13 at 17:36
  • That said, your proof is much cleaner, very nicely done. There is a strong connection between your {$n_i$} and my {$c_n$}. – Betty Mock Sep 17 '13 at 17:42
  • "It's about forcing $a_n<1/n^q$" that inequality is satisfied by $a_n=1/(2n^q)$, and $a_n<1/n$ is satisfied by $a_n=1/(2n)$, or if you prefer, $a_n=1/(n\log(n))$; I know you don't want to allow these, but I don't think you have really said what you mean. – Jonas Meyer Sep 17 '13 at 19:01
  • Thanks for the compliment. – Jonas Meyer Sep 17 '13 at 19:01
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    "$...a_nn^q$ cannot have infinitely many terms greater than 1 or the series $\sum 1/n^q$ where $q>1$ will not converge..." I do not see this. I see that $\sum_n a_nn^q/n$ must converge. This means that it is not possible that $a_nn^q\geq 1$ for all $n$, because the harmonic series diverges. However, the negation of that does not imply that $a_nn^q<1$ for all but finitely many $n$. How about $a_n=1/n^q$ when $n$ is a power of $2$, and $a_n=1/2^n$ otherwise. This would satisfy the criteria, but with $a_nn^q=1$ for infinitely many $n$. – Jonas Meyer Sep 17 '13 at 19:09
  • You have pointed out that I did not do this well, and I will revise it. Re your last note, if $a_n = 1/n^q$ when n is $2^k$ then that $a_n = 1/2^{kq}$. Since q is fixed eventually kq > 1, and $a_n$ becomes smaller than 1. – Betty Mock Sep 17 '13 at 20:30
  • @BettyMock: Yes, in that example $a_n\to0$, and in particular $a_n$ is eventually smaller than $1$; $\sum_na_n$ is a convergent series. But this doesn't help that infinitely often $a_nn^q=1$. – Jonas Meyer Sep 17 '13 at 20:47
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    Regarding the latest edit: It is still not true that we can conclude that $a_n<1/n$; you will have to quantify $n$ to turn that into a correct statement. Regardless it is impossible to conclude that $\sum _{2^{n-1}+1}^{2^n}a_n<1/2^n$. (Also $n$ is overloaded, and I cannot follow what leads up to that inequality, but the conclusion is false.) Note that an arbitrary convergent series of positive terms $\sum_n a_n$ satisfies the hypothesis, and a convergent series like $\sum_n 1/n^2$ does not satisfy this inequality. – Jonas Meyer Sep 18 '13 at 01:16
  • If your approach is to derive convergence from the (not always true) statement that for all $j\in\mathbb N$, $a_n<\dfrac{1}{j\cdot n}$ for sufficiently large $n$, then the counterexample $a_n=\dfrac{1}{n\log(n)}$ is appropriate, because $\sum_{n=2}^\infty \dfrac{1}{n\log(n)}$ diverges. – Jonas Meyer Sep 18 '13 at 01:21
  • Edited the proof to show that $a_nn^q$ cannot be > 1 for infinitely many n. I hope you think that part is all right. I was suspicious myself of the convergence part, and will revisit it when my work deadline is past. Thanks for your comments. – Betty Mock Sep 19 '13 at 17:22