1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.