# Baby Rudin ex. 3.8 proof verification

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.

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.