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I want to show that a series of the form $\sum a_n b_n$ converges where $\{a_n\}$ and $\{b_n\}$ satisfy

  • $a_n \ge 0$ and $\sum a_n$ converges
  • $b_0 \ge b_1 \ge b_2 \ge \cdots $ and $b_n \rightarrow b$ where $b$ need not be zero.

The case where $b = 0$ is easy. It is exactly same as theorem 3.42 (Dirichlet Test on Wikipedia) from Rudin's Principles of Mathematical Analysis, where he uses the partial summation formula.

seeker
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  • after posting the question I noticed that this is actually an exercise from Rudin. How embarrassing. see here – seeker Dec 22 '17 at 03:54
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    Couldn't you just use $\sum a_nb_n\leq \sum a_nb_0$ and the monotone convergence theorem? – JMoravitz Dec 22 '17 at 04:02
  • I suspect you do not want to assume that the $a_n $ are positive, since the result is trivial in that case. – Andrés E. Caicedo Dec 22 '17 at 04:54
  • @AndrésE.Caicedo thanks for pointing that out. When both $a_n, b_n$ are positive then $\sum a_n b_n$ is a monotone increasing series which is convergent. I removed the nonnegative condition on $b_n$ now. – seeker Dec 22 '17 at 05:01
  • @JMoravitz if both $a_n, b_n$ are positive, it is trivial. I updated the question now, $b_n$ need not be nonnegative, so your argument may not go through. – seeker Dec 22 '17 at 05:04
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    If $b \neq 0$, then since you've proved the $b = 0$ case, it is easy to convert this problem to the proved one by considering $\beta_n \colon = b_n - b$. Here you could separate a convergent part $b \sum a_n$ and prove the rest. P.S. This is known as the Abel test in some text. – xbh Dec 22 '17 at 05:34

1 Answers1

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When $b \neq 0$, the proof from Rudin breaks down. Below is an attempt at using the same technique.

If $A_n = \sum_{k=1}^{n} a_n$, then we can choose an $M$ such that $|A_n| \le M$ for all $n$. For a given $\varepsilon \ge 0$, we can find an integer $N$ such that $b_N \le b + \frac{\varepsilon}{2M}$, then we can have

$$ \begin{aligned} \left|\sum_{k=n}^{m} a_k b_k \right| &= \left| \sum_{k=n}^{m-1} A_{k}(b_{k} - b_{k-1}) + A_m b_m - A_{n-1}b_n\right| \\ &\le M \left| \sum_{k=n}^{m-1} (b_{k} - b_{k-1}) + b_m + b_n \right| \\ &= M 2b_n \\ &\le 2Mb_N \\ &\le 2Mb + \varepsilon \\ \end{aligned} $$

The idea is to get rid of the term involing $b$. We can rewrite $\sum a_n b_n$ as

$$ \sum a_n b_n = \sum a_n b + \sum a_n d_n $$

where $d_n$ satisfies the conditions from theorem 3.42 of rudin. Now the two series on the RHS of above equation are convergent. Therefore $\sum a_n b_n$ is convergent for a bounded monotonic sequence $\{b_n\}$ as well.

seeker
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