Suppose that $\sum\limits_{n=1}^{\infty}a_n$converges.Prove that if $b_n$ monotone increasing to infinity and $\sum\limits_{n=1}^{\infty}a_nb_n$converges,then $b_n\sum\limits_{k=n}^{\infty}a_k\to0(n\to\infty)$.I made the following attempts. Let $R_n=\sum\limits_{k=n}^{\infty}a_k$,so $\lim\limits_{n\to\infty}R_n=0$.then we know that $\{\frac{1}{b_n}\}$ monotonic decreasing and $\frac{1}{b_n}\to0(n\to\infty)$.So we have $$\sum\limits_{k=n}^{\infty}\dfrac{a_k}{b_k}=\sum\limits_{k=n}^{\infty}\dfrac{1}{b_k}(R_k-R_{k+1})\implies\left|\sum\limits_{k=n}^{\infty}\dfrac{a_k}{b_k}\right|\leqslant\sum\limits_{k=n}^{\infty}\dfrac{1}{b_k}|R_k-R_{k+1}|\leqslant\dfrac{1}{b_n}\sum\limits_{k=n}^{\infty}|R_k-R_{k+1}|\leqslant\dfrac{1}{b_n}R_n.$$ So we can get it $b_n\sum\limits_{k=n}^{\infty}\dfrac{a_k}{b_k}\to0(n\to\infty)$.I'd like to know how to proceed. Thank you.

  • $\begingroup$ Are the $a_n$ non-negative? Then $b_n \sum_{k=n}^\infty a_k \le \sum_{k=n}^\infty a_k b_k \to 0$. $\endgroup$
    – Martin R
    Mar 25 '21 at 12:28
  • $\begingroup$ @ Martin R There is no explanation in the title, so... $\endgroup$ Mar 25 '21 at 12:30
  • $\begingroup$ Check this: math.stackexchange.com/q/1164031/42969 $\endgroup$
    – Martin R
    Mar 25 '21 at 14:26
  • $\begingroup$ @Martin R God,How do you find this problem...I probably looked at it and it didn't feel right. I thought, if we can prove it$b_n\sum\limits_{k=1}^{n}a_n\to0$.Does that mean $b_n\sum\limits_{k=n}^{\infty}a_k\to0(n\to\infty)$ holds? $\endgroup$ Mar 25 '21 at 14:33
  • $\begingroup$ approach0.xyz/search/… $\endgroup$
    – Martin R
    Mar 25 '21 at 14:39