Baby Rudin ex. 3.8 proof verification

Question

The question asks: if $\sum a_n$ converges, $\{b_n\}$ is monotonic and bounded, prove that $\sum a_nb_n$ converges.

My proof goes as follows:

Let $\varepsilon>0$, and let $S_k$ denote $k$-th partial sum of $\sum a_nb_n$. Now, $\{b_n\}$ is monotonic and bounded, so it converges, say to $b$.

Since $\{b_n\}$ converges, we know that $\|b_n\| < \|b\|+1$, for large enough $n$. Also, $\sum a_n$ converges, thus $\|\sum_{j+1}^{k} a_n\| < \frac{\varepsilon}{\|b\|+1}$ for large enough $k,j$.

Therefore $\|S_k-S_j\| = \|\sum_{j+1}^{k} a_nb_n\| < (\|b\|+1)\|\sum_{j+1}^{k} a_n\|<(\|b\|+1)\times\frac{\varepsilon}{\|b\|+1}=\varepsilon,$ for large enough $k,j$.

Thus, $\sum a_nb_n$ is Cauchy, hence convergent.

Now, is my proof correct? Thanks for the help.

Answer 1

Your proof is not valid. The inequality $ |\sum\limits_{n=j+1}^{k}a_nb_n| < (|b|+1)|\sum\limits_{n=j+1}^{k}a_n|$ is not valid since $a_n$'s may be positive or negative. A correct proof is as follows:

Let $s_n=a_1+a_2+\cdots+a_n$. Then $\sum\limits_{n=j}^{j+k}a_nb_n=s_j(b_j-b_{j+1})+s_{j+1}(b_{j+1}-b_{j+2})+\cdots+ s_{j+k-1}(b_{j+k-1}-b_{j+k})+s_{j+k}b_{j+k}$. Since $(s_n)$ converges it is bounded. Hence we get $|\sum\limits_{n=j}^{j+k}a_nb_n|\leq M ((b_j-b_{j+1})+(b_{j+1}-b_{j+2})+\cdots+(b_{j+k-1}-b_{j+k})+|b_{j+k}|$ where $M=\sup_n |s_n|$. If $(b_n)$ is decreasing and positive this the above sum tends to $0$ and the roof is complete. If $(b_n)$ is decreasing and bounded we can add a constant to make it positive. If $(b_n)$ is increasing we can change $(b_n)$ to $(-b_n)$ to complete the proof.