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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.

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

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