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!

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    $\begingroup$ Sorry, fixed it I think. $b_n \geq 0$ and decreasing monotonically to $0$. $\endgroup$ – user93370 Sep 6 '13 at 2:47
  • $\begingroup$ The first thing to notice is that your condition does imply that a$_n$ goes to zero. To see this consider that $\sum 1/n$ $\endgroup$ – Betty Mock Sep 6 '13 at 3:37
  • $\begingroup$ 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. $\endgroup$ – Betty Mock Sep 6 '13 at 3:48
  • $\begingroup$ @BettyMock: I have edited your comment so it renders properly (as you might have expected, there was a missing $). $\endgroup$ – user642796 Sep 6 '13 at 4:26
  • $\begingroup$ 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. $\endgroup$ – Betty Mock Sep 7 '13 at 3:30

(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.


$\sum a_nb_n$ 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.

  • $\begingroup$ 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! $\endgroup$ – user642796 Sep 8 '13 at 5:14
  • $\begingroup$ @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. $\endgroup$ – Betty Mock Sep 9 '13 at 17:04
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    $\begingroup$ 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. $a_n=1/(2n)$. $\endgroup$ – Jonas Meyer Sep 17 '13 at 11:38
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    $\begingroup$ "$...a_nn^q$ 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 $n$. $\endgroup$ – Jonas Meyer Sep 17 '13 at 19:09
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    $\begingroup$ 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. $\endgroup$ – Jonas Meyer Sep 18 '13 at 1:16

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