If $\sum a_n$ converges and $\left\{b_n\right\}$ is monotonic and bounded, then $\sum a_n b_n$ converges.

Here's Prob. 8, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

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

My effort:

Let $$b = \lim_{n \to \infty} b_n$$.

Let's first suppose that $$\left\{ b_n \right\}$$ is monotonically decreasing. Then $$b_n \geq b$$ for all $$n$$. So we have $$b_0 - b \geq b_1 - b \geq \cdots \geq 0.$$ Moreover the sequence of the partial sums of $$\sum a_n$$ is bounded and $$\lim_{n \to \infty} \left( b_n - b \right) = 0$$.

So by Theorem 3.42 in Rudin, we can conclude that $$\sum a_n (b_n - b)$$ converges. And since $$a_n b_n = a_n (b_n - b) + a_n b$$ and since $$\sum a_n$$ converges, therefore $$\sum a_n b_n$$ converges by Theorems 3.47 in Rudin.

Now let's suppose that $$\left\{ b_n \right\}$$ is monotonically increasing. Then $$\left\{ -b_n \right\}$$ is monotonically decreasing and so $$\sum a_n \left(-b_n\right)$$ converges. Therefore by Theorem 3.47 in Rudin $$\sum a_n b_n$$ converges as well.

Is my proof correct? If not, then where is it wanting?

• I'm not sure what this theorem 3.42 is, but assuming you are using it correctly there's no mistake in your argument; also for the last argument I suppose convergence of $\sum a_n(-b_n)$ enough to say that $\sum a_nb_n$ converges as $\sum a_nb_n=-\sum a_n(-b_n)$ Dec 25 '16 at 15:18
• @user160738 Theorem 3.42 in Rudin reads as follows: Suppose the partial sums $A_n$ of $\sum a_n$ form a bounded sequence; $b_0 \geq b_1 \geq b_2 \geq \cdots$; and $\lim_{n \to \infty} b_n = 0$. Then $\sum a_n b_n$ converges. Dec 25 '16 at 17:08
• @user160738 can you please have a look at my post again and answer the question(s)? Dec 30 '16 at 17:46

Set $A_n=a_1+\cdots+a_n$, then $$\sum_{k=1}^na_kb_k=\sum_{k=1}^n (A_k-A_{k-1})b_k=\sum_{k=1}^nA_kb_k -\sum_{k=2}^nA_{k-1}b_k=\sum_{k=1}^nA_kb_k -\sum_{k=1}^{n-1}A_{k}b_{k+1}=\sum_{k=1}^{n-1}A_k(b_k-b_{k+1})+A_nb_n$$ Clearly, $A_nb_n$ converges as a product of two converging sequences, while $s_n=\sum_{k=1}^{n-1}A_k(b_k-b_{k+1})$, also converges, because it converges absolutely: $$\sum_{k=1}^{n-1}|A_k| |b_k-b_{k+1}|\le \big(\sup A_n\big)\cdot |b_1-b|,$$ where $\,b=\lim b_n$.