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Problem: If $\sum a_n$ converges, and $\{b_n\}$ is monotonic and bounded, prove that $\sum a_n b_n$ converges.

Source: Rudin, Principles of Mathematical Analysis, Chapter 3, Exercise 8.

Potato
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2 Answers2

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The sequence $\{b_n\}$ is monotonic and bounded, so it converges to some number $C$. Assume, without loss of generality, that the sequence $\{b_n\}$ is increasing, and write $b_n=C-d_n$, where $d_n\rightarrow 0$. We have

$$\sum a_nb_n = C\sum a_n -\sum a_nd_n.$$

The first series on the right is convergent by hypothesis, and the second is convergent because of the following theorem:

Theorem: If the partial sums of $\sum t_n$ form a bounded sequence and $s_n$ is a decreasing sequence that tends to 0, then $\sum t_ns_n$ converges.

Here we take $t_n=a_n$ and $s_n=d_n$.

Potato
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  • Did you just answer your own question! =) – Lemon Jun 26 '12 at 20:56
  • Since this is a rather old problem so I don't know if you are still around, @Potato: but could you tell me if this approach you did for this problem, especially the part $b_n = C - d_n$ a rather common technique to use ? – hyg17 Jun 05 '13 at 08:49
  • @hyg17 I don't think so, but it's been a while since I've done a lot of series manipulation. – Potato Jun 05 '13 at 13:02
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    Hi Potato, the theorem you mentioned has been just as difficult for me to prove as the original problem. For example, we have to worry about $\sum t_n$ diverging just as much as we have to worry about $\sum b_n$ diverging. Is there a simple proof to the result you mentioned? – Doug Aug 24 '13 at 06:07
  • @DanDouglas It's slightly tricky. You need to use summation by parts. The following 2 images are excerpts from Rudin's book, pages 70 and 71. Click here. Let me know if they answer your question or if you need more detail. – Potato Aug 24 '13 at 06:40
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I recently attempted this problem and came up with a solution which I don't think is as good as other answers here, but i want to know if it's correct or not. So here is my solution :

According to the hypothesis $\sum a_n$ converges and $\{b_n\}$ is bounded & monotonic. So there exists $T,M \gt 0$ such that $|A_n| = \left|\displaystyle\sum_{k=1}^{n} a_k\right| \le M$ and $|b_n| \le T$.

For every $\epsilon \gt 0$, there exists an $N_1 \gt 0$ such that $N_1 \le p \le q$ implies $\left|\displaystyle\sum_{n=p}^{q} a_n \right| \le \dfrac{\epsilon}{2T} \tag 1$

and again for every $\epsilon \gt 0$, there exists an $N_2 \gt 0$ such that $N_2 \le p \le q$ implies $|b_p - b_q| \le \dfrac{\epsilon}{4M} \tag2.$

$N = \text{max}(N_1, N_2)$

$\begin{align} \left|\displaystyle\sum_{n=p}^{q} c_n\right| = \left| \displaystyle\sum_{n=p}^{q} a_n b_n \right| &= \left| \displaystyle\sum_{n=p}^{q-1}A_n(b_n-b_{n+1}) + A_q b_q - A_{p-1} b_p\right| \\ &= \left| \displaystyle\sum_{n=p}^{q-1}A_n(b_n-b_{n+1}) + A_q(b_q - b_p) - (A_{p-1} - A_q)b_p \right| \\ &\le M\left| \displaystyle\sum_{n=p}^{q-1}(b_n-b_{n+1}) \right| + M\left| b_q - b_p \right| + T\left| \displaystyle\sum_{n=p}^{q} a_n \right| \\ &= 2M\left| b_q - b_p \right| + T\left| \displaystyle\sum_{n=p}^{q} a_n \right| \\ &\le 2M\left( \dfrac{\epsilon}{4M} \right) + T\left( \dfrac{\epsilon}{2T} \right) = \dfrac{\epsilon}{2} + \dfrac{\epsilon}{2} = \epsilon \end{align}$

So basically this mean that for every $\epsilon \gt 0$ there exists an $N \gt 0$ such that $N \le p \le q$ implies $\left| \displaystyle\sum_{n=p}^{q} a_n b_n \right| \le \epsilon.$

Soham Gadhave
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