# Too simple proof for convergence of $\sum_n a_n b_n$?

My question relates to Chapter 3, Exercise 8 in "Baby Rudin". It states:

If $$\sum_n a_n$$ converges, and if $$\{b_n\}$$ is monotonic and bounded, prove that $$\sum_n a_n b_n$$ converges.

My attempt would have been:

Since $$\{b_n\}$$ is monotonic and bounded, $$\{b_n\}$$ converges and it exists $$\inf \{b_n\}$$ as well as $$\sup \{b_n\}$$. But then we have $$\left| \sum_n a_n \inf \{b_n\} \right| \leq \left| \sum_n a_n b_n \right| \leq \left| \sum_n a_n \sup \{b_n\} \right| \leq \max \left( \left| \sup \{b_n\} \right|, \left| \inf \{b_n \} \right| \right)\varepsilon \leq \tilde{\varepsilon}$$ since $$\sum_n a_n$$ converges and $$\max \left( \left| \sup \{b_n\} \right|, \left| \inf \{b_n \} \right| \right)$$ is a finite number. This would imply, by the comparison test, that $$\sum_n a_n b_n$$ converges as well. $$\quad \Box$$

But as there was a way longer, more rigorous proof chosen in this solution manual, I'm a bit suspicious that my proof is not complete. Am I missing something?

EDIT: Thanks everyone! @hermes:

The last part of your proof gave me the following idea. As $$\lim_{n \to \infty} b_n = c$$ and $$\{b_n\}$$ is monotonic, couldn't we just set $$b_n = c - c_n$$ with a monotonically decreasing sequence $$\{c_n\}$$ which has $$\lim_{n \to \infty} c_n = 0$$. Then we have

$$\sum_n a_n b_n = \underbrace{c \sum_n a_n}_{\text{converges by assumption}} - \underbrace{\sum_n a_n c_n}_{\text{converges by Theorem 3.42}} \leq \varepsilon_1 - \varepsilon_2 = \varepsilon$$

since

Theorem 3.42 Suppose

• the partial sums of $$\sum_n a_n$$ form a bounded space $$\quad \checkmark$$
• $$c_0 \geq c_1 \geq c_2 \geq \dots \quad \checkmark$$
• $$\lim_{n \to \infty} c_n = 0 \quad \checkmark$$

Then $$\sum_n a_n c_n$$ converges.

Thus $$\sum_n a_n b_n$$ converges as well. $$\quad \Box$$

Now that should hold, I think. So I wouldn't need to go through all the estimates.

• Is the comparison test already treated in Chapter 3? Sep 13 '15 at 10:06
• If the $a_n$ don't all have the same sign, your inequalities don't necessarily hold. Sep 13 '15 at 10:07
• Yes, it is. As well as root and ratio test.
– root
Sep 13 '15 at 10:07
• "Solutions manual developed by Roger Cooke of the University of Vermont", it's not from Rudin himself. Sep 13 '15 at 10:10
• I would use Cauchy's convergence test, leveraging your remark that $(b_n)$ converges. Sep 13 '15 at 11:15

As Daniel Fischer points out, if the $a_n$ don't all have the same sign, your inequalities don't necessarily hold. You should use partial summation (used in proving Abel theorem) to get an estimate of Cauchy sum.
Let $A_n=\sum_{k=m}^n a_k$. So by Cauchy Criterion, $|A_n|<\epsilon$ for $n,m>N$.
We have \begin{align} \sum_{k=m}^n a_kb_k&=\sum_{k=m}^n (A_k-A_{k-1})b_k \\ &=\sum_{k=m}^n A_kb_k -\sum_{k=m}^n A_{k-1}b_k \\ &=\sum_{k=m}^{n-1} A_k(b_k-b_{k+1})+A_nb_n\tag{$A_{m−1}=0$} \end{align} First suppose $\lim_{n\to\infty}b_n=0$ and $\:b_n \downarrow$. Then $\:b_n\geqslant0$, and $b_k-b_{k+1}\geqslant0\:$ for all $k>0$.
Since $-\epsilon<A_k<\epsilon$ for all $k>m$ $$|A_k(b_k-b_{k+1})|<\epsilon(b_k-b_{k+1})$$ So for all $n,m>N-1$, there is \begin{align} \left|\sum_{k=m}^n a_kb_k\right|&\leqslant\sum_{k=m}^{n-1} |A_k(b_k-b_{k+1})|+|A_nb_n| \\ &\leqslant\sum_{k=m}^{n-1} \epsilon\:(b_k-b_{k+1})+\epsilon \:b_n \\ &=\epsilon \:(b_m-b_n+b_n) \\ &=\epsilon \:b_m \\ &\leqslant M\epsilon \end{align} So by Cauchy Criterion, $\sum_{k=1}^{\infty} a_kb_k$ converges.
Finally if $\lim_{n\to\infty}b_n=c\ne0$, we replace $b_n$ with $b_n-c$. Then $$\sum_{k=1}^{\infty} a_kb_k=\sum_{k=1}^{\infty} a_k(b_k-c)+c\sum_{k=1}^{\infty} a_k$$ converges for $\sum_{k=1}^{\infty} a_k(b_k-c)$ converges.